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1.7a + 3b = 11.1 1.2c + 0.3d = 3.9 and β0.2c β 0.4d = 0.4 ax + cy = 6 and 3ax β 2cy = β42 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The elimination method is usually more reliable than trying to solve complicated systems of equations graphically. Just be really careful with the signs when youβre subtracting. 248248248248248 Section 5.2 Section 5.2 Section 5.2 β The Elimination Method Section 5.2 Section 5.2 TTTTTopicopicopicopicopic 5.3.15.3.1 5.3.15.3.1 5.3.1 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques hniques aic tec aic tec hniques to aic techniques hniques aic tec solve rate problems, work problems, and percent mixture problems. What it means for you: Youβll solve real-life problems involving systems of linear equations. Key words: system of linear equations substitution method elimination method Check it out: 8.50c is the amount spent on CDs. 12.50d is the amount spent on DVDs. Section 5 |
.3 tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations Systems of Equa tions Equa Systems of tions Systems of tions tions pplica pplica β β β β β AAAAApplica tions pplications tions pplica tions tions pplica pplica β β β β β AAAAApplica tions pplications pplica tions Now itβs time to use all the systems of equations methods from Sections 5.1 and 5.2 to solve some real-life problems. e Pre Proboboboboblems lems lems e Pre Pr eal-Lif tions Help Solve Re Re Re Re Real-Lif eal-Lif tions Help Solv tions Help Solv Equa Equa Systems of Systems of lems eal-Life Pr Equations Help Solv Systems of Equa lems eal-Lif tions Help Solv Equa Systems of Systems of You can use systems of linear equations to represent loads of real-life problems. Once youβve written a system of equations, you can then use either substitution or elimination to solve the system of equations and therefore solve the problem. h Equationtiontiontiontion h Equa h Equa les in Eac ou Need the Same TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbles in Eac les in Eac ou Need the Same YYYYYou Need the Same ou Need the Same les in Each Equa h Equa les in Eac ou Need the Same Example Example Example Example Example 11111 A store has a year-end sale on CDs and DVDs. Each CD is reduced to $8.50, and the DVDs are going for $12.50 each. Akemi bought a total of 15 items (some CDs and some DVDs) for $163.50. Determine how many CDs and DVDs she bought. Solution Solution Solution Solution Solution One way to find the number of CDs and DVDs is to set up a system of two linear equations in two variables. Represent the unknown values with variables β Let c = number of CDs Akemi bought d = number of DVDs Akemi bought Since Akemi bought a total of 15 CDs and DVDs: c |
+ d = 15 β this is y st equationtiontiontiontion st equa st equa our fir our fir β this is y β this is y our first equa β this is your fir st equa our fir β this is y Now, consider the fact that each CD cost $8.50, each DVD cost $12.50, and Akemi spent $163.50. This leads to the equation: 8.50c + 12.50d = 163.50 β this is y our second equationtiontiontiontion our second equa our second equa β this is y β this is y β this is your second equa our second equa β this is y You now have a system of two linear equations in two variables, c and d. Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 249249249249249 Check it out This part is nothing new β it is the elimination method you used in Section 5.2. Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Now solve your system of equations: c + d = 15 8.50c + 12.50d = 163.50 Step 1: Make the coefficient of c the same in both equations. Do this by multiplying the first equation (c + d = 15) by 8.50. 8.50(c + d = 15) 8.50c + 8.50d = 127.50 Now you have two equations that have the same coefficient of c: 8.50c + 12.50d = 163.50 and 8.50c + 8.50d = 127.50 Step 2: Subtract one equation from the other to eliminate c. 8.50c + 12.50d = 163.50 β 8.50c + 8.50d = 127.50 4d = 36 ο¬ d = 9 Step 3: Substitute d = 9 into one of the original equations. Donβt forget Always remember to answer the original question clearly, in words. c + d = 15 c + 9 = 15 c = 6 So Akemi bought 6 CDs and 9 DVDs. Check the solution in the other equation: 8.50c + 12.50d = 163.50 8.50(6) + 12.50(9) = 163.50 163.50 = 163.50 β True Guided |
Practice 1. Pedro bought a total of 18 paperback and hardcover books for a total of $150. If each paperback was on sale for $6.50 and the hardcovers were on sale for $9.50 each, calculate how many paperbacks and how many hardcovers Pedro bought. 2. Three cans of tuna fish and four cans of corned beef cost $12.50. However, six cans of tuna fish and three cans of corned beef cost $15. How much does each type of can cost individually? 3. A school raised funds for its sports teams by selling tickets for a play. A ticket cost $5.75 for adults and $2.25 for students. If there were five times as many adults at the play than students and ticket sales raised $2480, how many adult and student tickets were sold? 4. Teresa bought five cups and four plates for $14.50. However, five plates and four cups would have cost $14.75. Find the cost of each item individually. 5. Simon has a total of 21 dimes and quarters in his coin bank. If the value of the coins is $4.05, how many coins of each type does Simon have? 250250250250250 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 tions to Solve e e e e AgAgAgAgAge Questions e Questions e Questions tions to Solv tions to Solv Equa Equa Using Systems of Using Systems of e Questions Equations to Solv Using Systems of Equa e Questions tions to Solv Equa Using Systems of Using Systems of Example Example Example Example Example 22222 The sum of Joseβs and Elizabethβs ages is 40 years. Five years ago, Elizabeth was four times as old as Jose. How old are Jose and Elizabeth now? Solution Solution Solution Solution Solution First form a system of two equations. Present Ages: Let x = Joseβs age y = Elizabethβs age The sum of their ages is 40 ο¬ x + y = 40 Ages 5 years ago: x β 5 = Joseβs age y β 5 = Elizabethβs age Elizabethβs age was four times Joseβs age ο¬ y β 5 = 4(x β 5) ο¬ y β 5 = 4x β 20 ο¬ y β 4x = β15 So the |
system of equations is: x + y = 40 y β 4x = β15 Now solve the system of equations. The y-coefficients are the same, so subtract one equation from the other to eliminate y: y + x = 40 β y β 4x = β15 5x = 55 ο¬ x = 11 Now substitute 11 for x in an original equation. x + y = 40 11 + y = 40 y = 29 Therefore Jose is 11 years old and Elizabeth is 29 years old. Check the solution in the other equation. y β 4x = β15 29 β 4(11) = β15 β15 = β15 β True Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 251251251251251 Guided Practice 6. The sum of Julioβs and Charlesβs ages is 44 years. Seven years from now, Julio will be twice Charlesβs age now. How old is each one now? 7. A mother is five times as old as her daughter. In five years, the mother will be three times as old as her daughter. How old is each now? 8. Anthony is twice as old as Teresa. In 20 years, the sum of their ages will be 43. How old is each now? 9. The sum of Ivy's and Audrey's ages is 27. Nine years ago, Ivy was twice as old as Audrey. How old is each now? 10. Three years from now, a father will be 10 times as old as his daughter. One year ago, the sum of their ages was 36. How old is each now? Independent Practice 1. Eight goldfish and 12 angelfish cost $53. Five goldfish and eight angelfish cost $34.75. If all goldfish are the same price and all angelfish are the same price, find the price of each goldfish and each angelfish. 2. Seven pairs of shorts and nine shirts cost $227.95. Three pairs of shorts and five shirts cost $116.55. If all pairs of shorts are the same price and all shirts are the same price, how much does each pair of shorts and each shirt cost? 3. Find the values of x and y if the figure shown on the right is a rectangle. 3x β y 2x + y 6 (5x + y) inches (7x β 5) inches (2y + 7 |
) inches 14 4. By finding the values of x and y, calculate the area of the rectangle on the left. All dimensions are in inches. (3x + 5) inches 5. Given that the triangle on the right is an isosceles triangle with a 40 cm perimeter, find the values of x and y 2x + 3y ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Systems of linear equations are really useful for working out real-life problems. In the rest of this Section youβll see more everyday situations modeled as systems of equations. 252252252252252 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 TTTTTopicopicopicopicopic 5.3.25.3.2 5.3.25.3.2 5.3.2 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques hniques aic tec aic tec hniques to aic techniques hniques aic tec solve rate problems, work problems, and percent mixture problems. What it means for you: Youβll solve integer problems involving systems of linear equations. Key words: system of linear equations substitution method tions tions Equa Equa Systems of |
Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa tions Equa Systems of Systems of lems lems er Proboboboboblems er Pr er Pr β Inteβ Inteβ Inteβ Inteβ Integgggger Pr lems lems er Pr er Proboboboboblems β Inteβ Inteβ Inteβ Inteβ Integgggger Pr lems lems er Pr er Pr lems er Pr lems In this Topic youβll solve systems of linear equations to figure out solutions to problems involving integers. er Proboboboboblems lems lems er Pr e Integgggger Pr er Pr e Inte e Inte tions to Solv tions to Solv Equa Equa Using Systems of Using Systems of lems tions to Solve Inte Equations to Solv Using Systems of Equa lems er Pr e Inte tions to Solv Equa Using Systems of Using Systems of Example Example Example Example Example 11111 The sum of two integers is 53. The larger number is 7 less than three times the smaller one. Find the numbers. Solution Solution Solution Solution Solution First form a system of two equations. Let x = smaller number y = larger number The sum of the two integers is 53, so x + y = 53. The larger is 7 less than 3 times the smaller one, so y = 3x β 7. So the system of equations is x + y = 53 and y = 3x β 7. Now solve the system of equations. The variable y in the second equation is already expressed in terms of x, so it makes sense to use the substitution method. Substitute 3x β 7 for y in the first equation. x + y = 53 x + (3x β 7) = 53 4x β 7 = 53 4x = 60 x = 15 Now, substitute 15 for x in the equation y = 3x β 7. y = 3x β 7 y = 3(15) β 7 y = 45 β 7 y = 38 That means that the integers are 15 and 38. The solution should work in both equations: x + y = 53 ο¬ 15 + 38 = 53 ο¬ 53 = 53 β True y = 3x β 7 ο¬ 38 = 3(15) β 7 οΏ½ |
οΏ½ 38 = 38 β True A useful check is to make sure that the answer matches the information given in the question: The sum of the two integers is 15 + 38 = 53. This matches the question. Three times the smaller integer is 15 Γ 3 = 45. The larger integer is 45 β 38 = 7 less than this. Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 253253253253253 Guided Practice 1. The sum of two numbers is 35 and the difference between the numbers is 13. Find the numbers. 2. The sum of two integers is 6 and the difference between the numbers is 40. Find the numbers. 3. The difference between two numbers is 50. Twice the higher number is equal to three times the lower number. Find the numbers. 4. The difference between two numbers is 11. If the sum of twice the higher number and three times the lower number is 137, find the numbers. 5. There are two numbers whose sum is 64. The larger number subtracted from four times the smaller number gives 31. Find the two numbers. Independent Practice 1. The length of a rectangle is three times the width. The perimeter is 24 inches. Find the length and the width. 2. In a soccer match, Team A defeated Team B by 3 goals. There were a total of 13 goals scored in the game. How many goals did each team score? 3. An outdoor adventure club has 35 members, who are all either rock climbers or skiers. There are 3 more rock climbers than there are skiers. How many climbers are there? 4. The tens digit of a two-digit number is three times the ones digit. If the sum of the digits is 12, find the two-digit number. 5. The sum of the digits of a two-digit number is 13. The number is 27 more than the number formed by interchanging the tens digit with the ones digit. Find the number. 6. Subtracting the ones digit from the tens digit of a positive two-digit number gives β2. Find the number given that the number is 18 less than the number formed by reversing its digits. (More than one answer is possible.) 7. The ones digit of a positive two-digit number is seven more than the tens digit. If the number formed by reversing the digits is 63 more than the original number, find the original number. (More than one answer is possible.) |
8. In a two-digit number, the sum of the digits is 10. Find the number if it is 36 more than the number formed by reversing its digits. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up For questions like this, it doesnβt matter which letters you choose to represent the unknown quantities. Itβs a good idea to pick letters that remind you in some way of what they represent. For example, in Example 1 involving CDs and DVDs, the letters c and d were sensible choices. 254254254254254 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 TTTTTopicopicopicopicopic 5.3.35.3.3 5.3.35.3.3 5.3.3 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques tototototo hniques hniques aic tec aic tec aic techniques hniques aic tec solvsolvsolvsolvsolveeeee rate problems, work problems, and perperperperpercent cent cent cent cent lems..... e pre proboboboboblems lems lems e pre pr mixtur mixtur mixture pr lems mixtur m |
ixtur What it means for you: Youβll solve percent mix problems involving systems of linear equations. Key words: percent mix system of linear equations Donβt forget: 5 5% is 100, so the number of gallons of real fruit juice in a gallons of apple drink is 0.05a. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of lems lems cent Mix Proboboboboblems cent Mix Pr cent Mix Pr β Pβ Pβ Pβ Pβ Pererererercent Mix Pr lems lems cent Mix Pr cent Mix Proboboboboblems β Pβ Pβ Pβ Pβ Pererererercent Mix Pr lems lems cent Mix Pr cent Mix Pr lems cent Mix Pr lems Percent mix problems are questions that involve two different amounts being mixed together to make a single mixture. tions tions Equa Equa lems Can Be Systems of cent Mix Proboboboboblems Can Be Systems of lems Can Be Systems of cent Mix Pr PPPPPererererercent Mix Pr cent Mix Pr tions Equations lems Can Be Systems of Equa tions Equa lems Can Be Systems of cent Mix Pr Example Example Example Example Example 11111 A high school sports coach decided to give a year-end party for all students at the school. The cafeteria manager decided to make 50 gallons of fruit juice drink by mixing some apple drink that contains 5% real fruit juice with strawberry drink that contains 25% real fruit juice. If the 50-gallon fruit juice drink is 10% real fruit juice, how many gallons of apple drink and strawberry drink did the cafeteria manager use? Solution Solution Solution Solution Solution First form a system of two equations. Let a = gallons of apple drink used s = gallons of strawberry drink used The total number of gallons is 50, so: a + s = 50 Now consider the total real fruit juice in the drink: (5% of a) + (25% of s) = 10% of 50 0.05a + 0.25s = 0.10(50) 0.05a + 0.25s = 5 5a + 25s = 500 a + 5s = 100 So the system of equations is: a + s = 50 |
a + 5s = 100 Now solve the system of equations. The coefficient of a is the same in both equations, so subtract one equation from the other to eliminate a. a + 5s = 100 β a + s = 50 4s = 50 ο¬ s = 12.5 Substitute 12.5 for s in one of the original equations. a + s = 50 a + 12.5 = 50 a = 37.5 That means that 37.5 gallons of apple drink was used and 12.5 gallons of strawberry drink. Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 255255255255255 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Check the solution in the other equation. a + 5s = 100 37.5 + 5(12.5) = 100 100 = 100 β True The answer must also match the information given in the question: 5% of 37.5 gallons of apple drink is real fruit juice. This is 0.05 Γ 37.5 = 1.875 gallons. 25% of 12.5 gallons of strawberry drink is real fruit juice. This is 0.25 Γ 12.5 = 3.125 gallons. The total fruit juice is 1.875 + 3.125 = 5 gallons. 5 gallons is 10% of the 50-gallon mixture, which matches the question. Guided Practice 1. A football coach calls a total of 50 plays in a game, some running plays and some passing plays. If 60% of the plays she calls are running plays, how many passing plays did she call? 2. During a fund-raiser, a girl collected a total of 60 quarters and halfdollars. If she raised $21.25, how many coins of each type did she collect? 3. A mother wants to make a quart of 2% fat milk for her children. She has a gallon of 1% fat milk and a gallon of whole milk (4% fat). How much of each should she mix to get a quart of 2% fat milk? 4. A ranger had 150 gallons of 60%-pure disinfectant that he mixed with 80%-pure disinfectant until the mixture was 70%-pure disinfectant. How much 80%-pure disinfectant was used by the ranger? 5. Two brands of tea are worth 60 cents per pound and 90 cents per pound respectively. How many pounds of each tea |
must be mixed to produce 138 pounds of a mixture that would be worth 80 cents per pound? 6. A pharmacist has a bottle of 10% boric acid and a bottle of 6% boric acid. A prescription requires 50 milliliters of a 7% boric acid solution. What volume of each solution should the pharmacist mix to get the desired solution? 7. In an industrial process, milk containing 4% butterfat is mixed with cream that contains 44% butterfat to obtain 250 gallons of 10% butterfat milk. How many gallons of 4% butterfat milk and 44% butterfat cream must be mixed to obtain the 250 gallons of 10% butterfat milk? 256256256256256 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 Independent Practice 1. A company invested a total of $5000, some at 6% and the rest at 8% per year. If the total return from the investments after one year was $330, how much money was invested at each rate? 2. Stacey used her 20% discount card to buy 7 cans of tomatoes and 15 cans of mushrooms for $10.64. The next day Brian spent $7.21 on 17 cans of tomatoes and 5 cans of mushrooms in a β30% off β sale. Find the undiscounted pre-sale price of each type of can. 3. Teresaβs piggy bank has a total of 100 dimes and nickels. If the piggy bank has a total of $7.50, how many of each coin does she have? 4. Anthony invested $1000 in two stocks. Stock A increased in value by 20%, while Stock B decreased in value by 10%. If Anthony ended the year with $1160 worth of stocks, how much money did he initially invest in each? 5. Dr. Baines makes a 5% metal hydride by mixing Material A, which contains 3% hydride, with Material K, which contains 7% hydride. If the final mixture weighs 10 kg, how much of each material does he mix? 6. Apples cost $0.75 per pound and pears cost $0.40 per pound. How much of each would need to be mixed to make 5 pounds of fruit salad worth $0.50 per pound? 7. Maddie invested $5000, some at a 5% rate of return and the rest at a |
2% rate of return per year. If the total return from the investments after one year was $200, how much money was invested at each rate? 8. Stephen collected a total of 75 quarters and half-dollars. If he collected $30.50, how many coins of each type did he collect? 9. Robert used his 10% discount card to buy 4 pizzas and 2 calzones for $45. The next day Audrey spent $15 on 1 pizza and 2 calzones in a β25% off β sale. Find the undiscounted price of pizzas and calzones. 10. Fred invested $6000, some at a 10% annual return and the rest at a 4% annual return. If the total return from the investments after one year was $400, how much money was invested at each rate? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Percent mix problems donβt have to involve liquids like the example in this Topic. You could mix together solids or even amounts of money β the solving method is always the same. Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 257257257257257 TTTTTopicopicopicopicopic 5.3.45.3.4 5.3.45.3.4 5.3.4 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15. |
0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques tototototo hniques hniques aic tec aic tec aic techniques hniques aic tec solvsolvsolvsolvsolveeeee rrrrraaaaate pr lems lems te proboboboboblems te pr te pr lems, work lems te pr problems, and percent mixture problems..... What it means for you: Youβll solve rate problems involving systems of linear equations. Key words: rate system of linear equations Check it out: When the boat travels downriver, itβs helped along by the water current. When the boat travels upstream, itβs slowed down by the water current. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of lems lems te Proboboboboblems te Pr te Pr β Rβ Rβ Rβ Rβ Raaaaate Pr lems lems te Pr te Proboboboboblems β Rβ Rβ Rβ Rβ Raaaaate Pr lems lems te Pr te Pr lems te Pr lems Hereβs the final application of systems of linear equations β rate problems deal with speed, distance, and time. tions TTTTToooooooooo tions tions Equa Equa lems Can Be Systems of te Proboboboboblems Can Be Systems of lems Can Be Systems of te Pr RRRRRaaaaate Pr te Pr Equations lems Can Be Systems of Equa tions Equa lems Can Be Systems of te Pr Example Example Example Example Example 11111 During a storm, a flood rescue team in a boat takes 3 hours to travel downriver along a 120-mile section of Nastie river. If it takes 4 hours to travel the same river section upriver, find the boatβs speed in still water and the speed of the water current. [Assume the boat has the same speed relative to the water and that the speed of the current remains constant.] Solution Solution Solution Solution Solution First form a system of two equations. Let x = boatβs speed in still water y = water current |
speed Downriver: When traveling downriver, the water speed adds to the boat speed. So boat speed downriver = x + y Use the formula Distance = Speed Γ Time: 120 miles = (x + y) Γ 3 hours 120 = 3(x + y) 120 3 = x + y 40 = x + y Upriver: When traveling upriver, the water speed acts against the boat speed. So boat speed upriver = x β y Use Distance = Speed Γ Time: 120 miles = (x β y) Γ 4 hours 120 = 4(x β y) = x β y 120 4 30 = x β y So the system of equations is: 40 = x + y 30 = x β y 258258258258258 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Now solve the system of equations. The y-coefficients have opposite values, so add the equations to eliminate y. 40 = x + y + 30 = x β y 70 = 2x ο¬ x = 35 Now substitute 35 for x in an original equation. 40 = x + y 40 = 35 + y y = 5 So, the boatβs speed in still water (x) is 35 mph and the water current speed (y) is 5 mph. Check it out: The distance was measured in miles and the time was measured in hours, so the units must be miles per hour (mph). Check the solution in the other equation. 30 = x β y 30 = 35 β 5 30 = 30 β True Guided Practice 1. The Lee High School crew team takes 2 hours to row downriver along a 60-mile section of the Potomac River. It takes 4 hours to travel the same river section upriver. Find the boatβs speed in still water and the speed of the water current. 2. A plane travels 300 miles in 40 minutes against the wind. Flying with the wind, the same plane travels 200 miles in 20 minutes. Find the planeβs speed without wind and the wind speed. 3. Sarah rides her bicycle to work, a journey of five miles (one way). One day the journey to work is against the wind, and takes 30 minutes. The ride back, with the wind, takes 20 minutes. Find the wind speed, and work out how long the ride (one |
way) would take with no wind. 4. A red car and a blue car start at the same time from towns that are 16 miles apart, and travel towards each other. The red car is 7 mph faster than the blue car. After 15 minutes the cars are 7 miles apart. Find the speed of each car. Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 259259259259259 Independent Practice 1. It takes a canoeist 1.5 hours to paddle down a 10-mile stretch of river. If it takes 3 hours to travel the same river section upriver, find the canoeistβs speed in still water and the speed of the water current. Round answers to the nearest tenth. 2. A medical helicopter pilot flies 200 miles with a tailwind in 2 hours. On the return trip, it takes 2.5 hours to fly against the wind. Find the speed of the helicopter in no wind and the speed of the wind. 3. A small powerboat travels down a 24-mile stretch of river in one hour. If it takes 3 hours to travel the same river section upriver, find the powerboatβs speed in still water and the speed of the water current. 4. A plane flies 700 miles with a tailwind in 5 hours. On the return trip, it takes 6 hours to fly the same 700 miles against the wind. Find the speed of the plane in no wind and the speed of the wind. Round answers to the nearest tenth. 5. Two planes start at the same time from cities that are 2500 miles apart, and travel toward each other. The rate of one plane exceeds the rate of the other by 15 mph. After 4 hours the airplanes were 1000 miles apart. Find the rate of each airplane. 6. Kyle lives in Washington, DC and Robert lives in Boston. The cities are 440 miles apart. They start driving toward each other at the same time. Kyle is driving an average of 5 mph faster than Robert. After 5 hours the cars are 180 miles apart. Find the rate of each car. 7. Sheri and Peter start running towards each other at the same time, from their houses that are 4000 feet apart. Sheri is running 2 feet per minute faster than Peter. After 5 minutes, they are still 300 feet apart. Find the running speeds of Sheri and Peter. 8. Casey leaves home at 10 a.m. and drives at an average speed |
of 25 mph. Marshall leaves the same house 15 minutes later, and drives the same route, but twice as fast as Casey. At what time will Marshall pass Casey, and how far will they be from home when he does? 9. Pittsburgh is 470 miles from Chicago and 350 miles from Philadelphia. Trains leave Chicago and Philadelphia at the same time, but the Chicago train travels 40 mph faster than the Philadelphia one. Both trains reach Pittsburgh at the same time. Find each trainβs speed, rounding answers to the nearest tenth. 10. At 07:00 a train leaves a station, traveling at 55 mph. At 07:30 an express train leaves the same station, traveling the same route at 70 mph. How long will it take the express train to overtake the other train, and how far will they be from the station when it does? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The word problems in this Section all look quite different β but you use the same methods each time to solve them. If youβre having trouble with solving them, look back at Sections 5.1 and 5.2 and try going through some examples that donβt involve real-life situations. 260260260260260 Section 5.3 Section 5.3 Section 5.3 β Applications of Systems of Equations Section 5.3 Section 5.3 Chapter 5 Investigation g Business g Business en in an Eg en in an Eg eaking Ev eaking Ev BrBrBrBrBreaking Ev g Business en in an Egg Business eaking Even in an Eg g Business en in an Eg eaking Ev BrBrBrBrBreaking Ev g Business g Business en in an Eg en in an Eg eaking Ev eaking Ev en in an Egg Business eaking Even in an Eg g Business en in an Eg eaking Ev g Business Even though this Investigation looks tough because itβs about money, itβs just systems of equations. Part 1: You decide to set up an egg-selling business. You plan to keep 12 hens in your yard and sell the eggs from your home. Free-Range Eggs 20Β’ each The initial costs of setting up the business are $47 for a hen house and $3 for each hen. The hens will cost a total of 50 cents a day to feed and you think each hen will lay one egg per day. You plan to sell the |
eggs for 20 cents each. 1) How many days will it be before you break even? Assume you manage to sell all the eggs laid without any going to waste. 2) On one set of axes, draw graphs to show how the costs and amount earned will change over the first 50 days. The βbreak-evenβ point is when the amount of money you have earned equals the amount of money youβve spent on the business. Part 2: After further research, you discover that this type of hen only lays eggs on four days out of five. However, you can buy special hens for $5 each that will not only lay one egg per day, but will lay two eggs every fourth day. You have a total of $98 to invest in setting up your business. How many of each type of hen should you buy to maximize your profits? Remember β the hen house will only accommodate up to 12 hens. Extension 1) Your mother is not happy about you using the yard for your business and demands 10% of your profit. With this in mind, calculate how much money you will make in the first year. 2) You are given $50 for your birthday. Will you make more money over one year if you buy another hen house and one hen or invest it at 6%? Investigate how receiving a different amount of money would affect the best option. Open-ended Extension Make up a business and describe its costs and the amount it charges for its products or services. Calculate when you will break even. Exchange business plans with a partner and each devise some changes that will affect the other personβs business. Investigate how your partnerβs changes will affect your profits and break-even point. Your break-even point doesnβt necessarily have to be after a certain number of days β it might be after you sell a milkshake ce ce ce ce company cccccererererertain numb ompany ompany milkshak tain numbererererer of your products. For example, a milkshak milkshak tain numb tain numb ompany might have an initial cost of buying a milkshak ompany tain numb edients edients ingr ingr blender and cups. There will then be the cost of the ingr edients for each milkshake made. After you sell a certain ingredients edients ingr equal equal equal to the amount that youβve spent so far |
. number of milkshakes you will have earned an amount equal equal ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you come across any real-life situation that involves two or more equations, youβll probably have to solve them using systems of equations β using all the skills you learned in this Chapter. estigaaaaationtiontiontiontion β Breaking Even in an Egg Business estigestig estig pter 5 Invvvvvestig pter 5 In ChaChaChaChaChapter 5 In pter 5 In pter 5 In 261261261261261 Chapter 6 Manipulating Polynomials Section 6.1 Adding and Subtracting Polynomials................. 263 Section 6.2 Multiplying Polynomials..................................... 274 Section 6.3 Dividing Polynomials......................................... 283 Section 6.4 Special Products of Binomials.......................... 297 Section 6.5 Factors.............................................................. 302 Section 6.6 Factoring Quadratics......................................... 310 Section 6.7 More on Factoring Polynomials......................... 318 Section 6.8 More on Quadratics........................................... 325 Investigation Pascalβs Triangle............................................... 331 262262262262262 Topic 6.1.1 Section 6.1 Polynomials Polynomials California Standards: 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll learn what polynomials are, and youβll simplify them. Key words: monomial polynomial like terms degree Check it out: The word monomial comes from the Greek words: mono β meaning one nomos β meaning part or portion. So monomial means one term. Check it out: The word polynomial comes from the Greek words: poly β meaning many nomos β meaning part or portion. So polynomial means many terms. Polynomial β another math word that sounds a lot harder than it actually is. Read on and youβll see that actually polynomials are not as complicated as you might think. A Monomial is a Single Term A monomial is a single-term expression. It can be either a number or a product of a number and one or |
more variables. For example, 13, 2xΒ², and βx3yn4 are all monomials. A Polynomial Can Have More Than One Term A polynomial is an algebraic expression that has one or more terms (each of which is a monomial). For example, x + 1 and β3xΒ² + 2x + 1 are polynomials. There are a couple of special types of polynomial: A binomial is a two-term polynomial, such as xΒ² + 1. A trinomial is a polynomial with three terms, such as β3xΒ² + 2x + 1. Guided Practice For each of the polynomials below, state whether it is a monomial, a binomial, or a trinomial. 1. 5x 3. 2y 5. 2y + 2 7. 7xΒ³ 9. xΒ² + 2x + 3 11. 4.3x β 8.9x2 + 4.2x3 13. 0.3x2y + xy2 + 4xy 15. a2 + b2 2. 9xΒ² + 4 4. 2xΒ²y 6. 5x β 2 8. xΒ² + y 10. 3x2 + 4x β 8 12. 0.3x2y 14. 8.7 16. 97.9a β 14.2c Section 6.1 β Adding and Subtracting Polynomials 263 Use Like Terms to Simplify Polynomials Like terms are terms that have exactly the same variables β for example, β2xΒ² and 5xΒ² are like terms. Like terms always have the same variables, but may have different coefficients. A polynomial can often be simplified by combining all like terms. Example 1 Simplify the expression 2xΒ² + 4y + 3xΒ². Solution Notice that there are two like terms, 2xΒ² and 3xΒ²: You can combine the like terms: 2xΒ² + 3xΒ² = 5xΒ² So 2xΒ² + 4y + 3xΒ² = 5xΒ² + 4y Guided Practice Simplify each of the following polynomials. 17. x + 1 + 2x 19. 9xΒ² + 4x + 7x 18. 3y + 2y 20. xΒ² + x + xΒ² Simplify each of the following polynomials, then state whether your answer is a mon |
omial, binomial, or trinomial. 21. 3x2 + 4 β 8 + x2 23. 3x2y β 2x2y + 8 25. 4x3 + 7 β x3 + 4 β 3x3 β 11 27. 3xy + 4xy + 5x2y β 4xy2 22. 8x3 + x4 β 6x3 + 4 24. 7 β 2y + 3 β 10 26. 5x2 + 9x2 + 4 + 2y 28. 9x5 + 2x2 + 4x4 + 5x5 β 3x4 β x2 Finding the Degree of a Polynomial The degree of a polynomial in x is the size of the highest power of x in the expression. If you see the phrase βa fourth-degree polynomial in x,β you know that it will contain at least one term with x4, but it wonβt contain any higher powers of x than 4. For example: has degree 1 2x + 1 yΒ² + y β 3 has degree 2 x4 β xΒ² has degree 4 β itβs a first-degree polynomial in x β itβs a second-degree polynomial in y β itβs a fourth-degree polynomial in x 264 Section 6.1 β Adding and Subtracting Polynomials Guided Practice State the degree of each of the following polynomials. 29. 3x + 5 31. xΒ² + 2x + 3 30. 3x4 + 2 32. x + 4 + 2xΒ² Simplify and state the degree of each of the following polynomials. 33. 2x + x2 + x β 3 34. 3a3 + 4a β 2a3 + 4a2 35. 4x3 + 4x8 β 3x8 + 2x3 36. 3y + 2y β 5y2 + 6y 37. b13 + 2b13 β 8 + 4 β 3b13 38. z3 + z3 β z6 + z7 + 3z7 39. c4 + c3 + c3 β c4 + c β 2c3 40. x β 2x9 β 8x4 + 13x2 Independent Practice For the polynomials below state whether they are a monomial, a binomial, or a trinomial. 1. 19 |
a2 + 16 3. 42xy 2. 2c β 4a + 6 4. 16a2b + 4ab2 Simplify each of the following polynomials. 5. 0.7x2 + 9.8 β x2 6. 17x2 β 14x9 + 7x9 β 7x2 + 7x9 7. 0.8x4 + 0.3x2 + 9.6 β x2 β 9x4 + 1.6x2 State the degree of the following polynomials. 8. x β 9x6 + 4 9. 14x8 + 16x10 + 4x8 10. 2x2 β 4x4 + 7x5 11. 2x2 β 4x + 8 Simplify each polynomial, state the degree of the polynomial, and determine whether it is a monomial, a binomial, or a trinomial. 12. 93a2 + 169 β 4a β 81a2 + 7 13. 7.9x2 β 13x4 β 1.5x4 + 1.4x2 1 15. 16. 1 9 14. 5x9 β 6x9 + 4 + x9 β 3 β 1 2 x9 β x3 β x3 + 1 x6 β 3 x10 β 3 5 4 x9 + 7 9 x10 β 3 4 17. When a third degree monomial is added to a second degree binomial, what is the result? 3 x6 + 1 9 18. When a 4th degree monomial is added to a 6th degree binomial, what are the possible results? Round Up Round Up This Topic gets you started on manipulating polynomials, by simplifying them. In the next couple of Topics youβll see how to add and subtract polynomials. Section 6.1 β Adding and Subtracting Polynomials 265 Topic 6.1.2 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll add polynomials and multiply a polynomial by a number |
. Key words: polynomial like terms inverse Adding Polynomials Adding Polynomials Adding polynomials isnβt difficult at all. The only problem is that you can only add certain parts of each polynomial together. The Opposite of a Polynomial The opposite of a number is its additive inverse. The opposite of a positive number is its corresponding negative number, and vice versa. For example, β1 is the opposite of 1, and 1 is the opposite of β1. To find the opposite of a polynomial, you make the positive terms negative and the negative terms positive. Example 1 Find the opposites of the following polynomials: a) 2x β 1 b) β5xΒ² + 3x β 1 Solution a) β2x + 1 b) 5xΒ² β 3x + 1 Donβt forget: See Topic 1.2.2 for more on additive inverses. Guided Practice Find the opposites of the following polynomials. 1. 2x + 1 3. xΒ² + 5x β 2 5. 3x2 + 4x β 8 7. 4x4 β 16 9. 5x4 β 6x2 + 7 11. β0.9x3 β 0.8x2 β 0.4x β 1.0 2. β5x β 1 4. 3xΒ² β 2x + 3 6. β8x2 β 4x + 4 8. 8x3 β 6x2 + 6x β 8 10. β2x4 + 3x3 β 2x2 12. β1.4x3 β 0.8x2 β 1 2 x 266 Section 6.1 β Adding and Subtracting Polynomials Adding Polynomials Adding polynomials consists of combining all like terms. There are a few ways of adding polynomials β two of the methods are shown in Example 2. Example 2 Find the sum of β5xΒ² + 3x β 1, 6xΒ² β x + 3, and 5x β 7. Solution Method A β Collecting Like Terms and Simplifying (β5xΒ² + 3x β 1) + (6xΒ² β x + 3) + (5x β 7) = β5xΒ² + 3x β 1 + 6xΒ² β x + 3 + 5x β 7 = β5xΒ² + 6xΒ² + 3x β x + 5 |
x β 1 + 3 β 7 = xΒ² + 7x β 5 Method B β Vertical Lining Up of Terms β5xΒ² + 3x β 1 + 6xΒ² β x + 3 + 5x β 7 xΒ² + 7x β 5 Both methods give the same solution. Donβt forget: See Topic 1.2.4 for the definition of multiplication. Multiplying a Polynomial by a Number Multiplying a polynomial by a number is the same as adding the polynomial together several times. Example 3 Multiply x + 3 by 3. Solution (x + 3) Γ 3 = (x + 3) + (x + 3) + (x + 3 = 3x + 9 Donβt forget: See Section 1.2 for more about the distributive property. The simple way to do this is to multiply each term of the polynomial by the number. In other words, you multiply out the parentheses, using the distributive property of multiplication over addition. Section 6.1 β Adding and Subtracting Polynomials 267 Example 4 Multiply xΒ² + 2x β 4 by 3. Solution 3(xΒ² + 2x β 4) = (3 Γ xΒ²) + (3 Γ 2x) β (3 Γ 4) = 3xΒ² + 6x β 12 Guided Practice Add these polynomials and simplify the resulting expressions. 13. (4x2 β 2x β 1) + (3x2 + x β 10) 14. (11x4 β 5x3 β 2x) + (β7x4 + 3x3 + 5x β 3) 15. β5c3 β 3c2 + 2c + 1 4c2 β c β 3 6c3 + c + 4 16. 5x2 + 3x β 3 β4x2 β 3x + 5 β2x2 + x β 7 Multiply these polynomials by 4. 17. 10y2 β 7y + 5 19. (βxΒ² + x β 4) 18. (xΒ² β 3x + 3) 20. (2xΒ² + 5x + 2) Independent Practice In Exercises 1-7, simplify the expression and state the degree of the resulting polynomial. 1. (2x2 + 3x β 7) + (7x2 β 3x + 4) 2. (x |
3 + x β 4) + (x3 β 8) + (4x3 β 3x β 1) 3. (βx6 + x β 5) + (2x6 β 4x β 6) + (β2x6 + 2x β 4) 4. (3x2 β 2x + 7) + (4x2 + 6x β 8) + (β5x2 + 4x β 5) 5. (0.4x3 β 1.1) + (0.3x3 + x β 1.0) + (1.1x3 + 2.1x β 2.0) 6. 7. β 4a3 β 2a + 3 8a4 β 2a3 β 4a + 8 7a4 β 4a β 7 1.1c2 + 1.4c β 0.48 β4.9c2 β 3.6c + 0.98 7.3c2 + 0.13 Multiply each polynomial below by β4. 8. 4a2 + 3a β 2 10. β6x3 β 4x2 + x β 8 Multiply each polynomial below by 2a. 12. 3a2 + a β 8 9. βc2 + 3c + 1 11. 24x3 + 16x2 β 4x + 32 13. β7a4 + 2a2 β 5a + 4 Round Up Round Up Adding polynomials can look hard because there can be several terms in each polynomial. The important thing is to combine each set of like terms, step by step. 268 Section 6.1 β Adding and Subtracting Polynomials Topic 6.1.3 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll learn how to subtract polynomials. Key words: polynomial like terms inverse Subtracting Polynomials Subtracting Polynomials Subtracting one polynomial from another follows the same rules as adding polynomials β you |
just need to combine like terms, then carry out all the subtractions to simplify the expression. Subtracting Polynomials Subtracting polynomials is the same as subtracting numbers. To subtract Polynomial A from Polynomial B, you need to subtract each term of Polynomial A from Polynomial B. Then you can combine any like terms to simplify the expression. Example 1 Subtract Polynomial A from Polynomial B, where Polynomial A = xΒ² + x and Polynomial B = xΒ² + 4x. Solution Subtract each term of Polynomial A from Polynomial B: Polynomial B β Polynomial A = xΒ² + 4x β (xΒ²) β (x) = xΒ² β xΒ² + 4x β x = 0 + 3x = 3x Guided Practice 1. Subtract x2 β 4 from x2 + 8. 2. Subtract 3x β 4 from 8x2 β 5x + 4. 3. Subtract x + 4 from x2 β x. 4. Subtract x2 β 16 from x2 + 8. 5. Subtract x2 + x β 1 from x + 4. 6. Subtract β3x2 + 4x β 5 from x2 β 7. 7. Subtract β3x2 β 5x + 2 from β2x3 β x2 β 7x. Simplify: 8. (9a β 10) β (5a + 2) 9. (5a2 β 2a + 3) β (3a + 5) 10. (x3 + 5x2 β x) β (x2 + x) Section 6.1 β Adding and Subtracting Polynomials 269 Donβt forget: See Topic 1.2.5 for the definition of subtraction. Subtracting is Simply Adding the Opposite Another way to look at subtraction of polynomials is to go back to the definition of subtraction. When you subtract Polynomial A from Polynomial B, what youβre actually doing is adding the opposite of Polynomial A to Polynomial B. Example 2 Subtract β5xΒ² + 3x β 8 from β7xΒ² + x + 5. Solution β7xΒ² + x + 5 β (β5xΒ² + 3x β 8 |
) = β7xΒ² + x + 5 + 5xΒ² β 3x + 8 = β7xΒ² + 5xΒ² + x β 3x + 5 + 8 = β2xΒ² β 2x + 13 Alternatively, you can do subtraction by lining up terms vertically: Example 3 Subtract β5xΒ² + 3x β 8 from β7xΒ² + x + 5. OR β7xΒ² + x + 5 + (5xΒ² β 3x + 8) β2xΒ² β 2x + 13 This is the opposite of β5xΒ² + 3x β 8 Solution β7xΒ² + x + 5 β (β5xΒ² + 3x β 8) β2xΒ² β 2x + 13 Guided Practice Simplify the expressions in Exercises 11β16. 11. (3a4 + 4) β (2a2 β 5a4) 12. (6x2 + 8 β 9x4) β (3x β 4 + x3) 13. (9c2 + 11c2 + 5c β 5) β (β10 + 4c4 β 8c + 3c2) 14. (8a2 β 2a + 5a) β (9a2 + 2a + 4) 15. 6x2 β 6 β(5x2 + 9) 16. 8a2 + 4a β 9 β(3a2 β 3a + 7) 17. Subtract 7a3 + 3a β 12 from 5a2 β a + 4 by adding the opposite expression. Use the vertical lining up method. 18. Subtract (8p3 β 11p2 β 3p) from 4p3 + 6p2 β 10 by adding the opposite expression. Use the vertical lining up method. 270 Section 6.1 β Adding and Subtracting Polynomials If you add a polynomial to its opposite, the result will always be 0. Example 4 Find the sum of β5xΒ² + 3x β 1 and 5xΒ² β 3x + 1. Check it out: You can use this method to check answers to βoppositesβ questions. Solution β5xΒ² + 3x β 1 + (5xΒ² β 3x + 1) = β5xΒ² + 3x β 1 + 5xΒ² β 3x + 1 = β5xΒ² + 5xΒ² + 3 |
x β 3x β Independent Practice Subtract the polynomials and simplify the resulting expression. 1. (5a + 8) β (3a + 2) 2. (8x β 2y) β (8x + 4y) 3. (β4x2 + 7x β 3) β (2x2 β 4x + 6) 4. (3a2 + 2a + 6) β (2a2 + a + 3) 5. β3x4 β 2x3 + 4x β 1 β (β2x4 β x3 + 3x2 β 5x + 3) 6. 5 β [(2k + 3) β (3k + 1)] 7. β 10a2 + 4a β 1 β (7a2 + 4a) 8. (x2 + 4x + 6) β (2x2 + 2x + 4) Solve these by first simplifying the left side of the equations. 9. (2x + 3) β (x β 7) = 40 11. (2 β 3x) β (7 β 2x) = 23 10. (4x + 14) β (β10x β 3) = 73 12. (17 β 5x) β (4 β 3x) β (6 β x) = 19 Find the opposite of the polynomials below. 13. x2 + 2x + 1 15. 4b2 β 6bc + 7c 14. βa2 + 6a + 4 16. a3 + 4a2 + 3a β 2 17. The opposite of a fifth degree polynomial has what degree? 18. If a monomial is subtracted from another monomial, what are the possible results? 19. What is the degree of the polynomial formed when a 2nd degree polynomial is subtracted from a 1st degree polynomial? 20. A 3rd degree polynomial has a 2nd degree polynomial subtracted from it. What is the degree of the resulting polynomial? Round Up Round Up Watch out for the signs when youβre subtracting polynomials. Itβs usually a good idea to put parentheses around the polynomial youβre subtracting, to make it easier to keep track of the signs. Section 6.1 β Adding and Subtracting Polynomials 271 Topic 6.1.4 California |
Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll use addition and subtraction of polynomials to solve more complicated problems. Key words: polynomial like terms Donβt forget: Remember that the perimeter of a rectangle is the sum of the lengths of all four sides. Donβt forget: Be really careful in part b) β the minus sign applies to both terms inside the parentheses. Adding and Subtracting Adding and Subtracting Polynomials Polynomials This Topic just contains one big example that uses both addition and subtraction of polynomials. An Example Using Addition and Subtraction Example 1 Find a) the sum of and b) the difference between the perimeters of the rectangles shown below 2xΒ² β 3 x xΒ² + 2 + 1 P 2 xΒ² β 1 Solution The first thing you need to do is find and simplify expressions for each of the perimeters. Perimeter of P1 = (3xΒ² β 3x β 2) + (3xΒ² β 3x β 2) + (2xΒ² β 3) + (2xΒ² β 3) = 3xΒ² β 3x β 2 + 3xΒ² β 3x β 2 + 2xΒ² β 3 + 2xΒ² β 3 = 3xΒ² + 3xΒ² + 2xΒ² + 2xΒ² β 3x β 3x β 2 β 2 β 3 β 3 = 10xΒ² β 6x β 10 Perimeter of P2 = (xΒ² + 2x + 1) + (xΒ² + 2x + 1) + (xΒ² β 1) + (xΒ² β 1) = xΒ² + 2x + 1 + xΒ² + 2x + 1 + xΒ² β 1 + xΒ² β 1 = xΒ² + xΒ² + xΒ² + xΒ² + 2x + 2x + 1 + 1 β 1 β 1 = 4xΒ² + 4x a) Perimeter of P1 + Perimeter of P2 = (10xΒ² β 6x β 10) + (4xΒ² |
+ 4x) = 10xΒ² β 6x β 10 + 4xΒ² + 4x = 10xΒ² + 4xΒ² β 6x + 4x β 10 = 14xΒ² β 2x β 10 b) Perimeter of P1 β Perimeter of P2 = (10xΒ² β 6x β 10) β (4xΒ² + 4x) = 10xΒ² β 6x β 10 β 4xΒ² β 4x = 10xΒ² β 4xΒ² β 6x β 4x β 10 = 6xΒ² β 10x β 10 272 Section 6.1 β Adding and Subtracting Polynomials Guided Practice Find the perimeter of each of these shapes. 2 4(5 β 3 + 1) x x 1. 2. 2 x 5( β x + 3) 3. 2 + 4 x x2 6( + ) x 3( β 1) x 2( + 7) x x2 7( + ) x Independent Practice Simplify each of these expressions. 1. 4y β [3(y β x) β 5(x + y)] 2. 3[2y β 3(y β 1) + 2(βy β 1)] 3. 3(x β y) β 2(x β 3y) 4. β2[(x β y) β 2(x + 2y)] β [(β2x β y) β (x + 2y)] 5. (β3x3 β x2y + 2xy2 β 2y3) β (β5x3 + x2y β 3xy2 β 3y3) + (βx3 β 2x2 y + y3) 3 β 3x 2) β x 3( 2 ( 7 β x ) 1) x β 3(2 2 ( 9 β x ) 6. Find the difference between the perimeters of the trapezoid and the triangle shown. 3(2 x β 1) 5 β 1x 7. Simplify 5(4x + 3) β 4(3x β 1) + 3(x β 1). 8. Simplify (8x3 β 5x2 β 2x β 7) β (6x3 β 3x2 + x β 5) β (β3x2 + 3x β 2). 9. Simplify β2(3x3 β 2x) β 3(βx3 + 7) β (2x2 β |
5x3 β 5). 10. Find the sum of the opposites of: β2x3 + 3x2 β 5x + 1 and 3x3 β 2x2 + 3x β 3 11. Find the difference between the opposites of: β2x2 β 3x + 5 and 3x2 + 2x β 4 12. Simplify β2(3x2p β 2xp + 1) + 4(2x2p β xp β 3) β 5(x2p β 2xp β 1). 13. If the perimeter of the figure shown is 90 inches, what are the dimensions of the figure? 2 ( x β 3 ) x 3(2 + 1) Round Up Round Up Youβve covered addition and subtraction of polynomials in just four short Topics. As you might expect, the next Section covers multiplication of polynomials, and then youβll learn about division of polynomials in Section 6.3. Section 6.1 β Adding and Subtracting Polynomials 273 Topic 6.2.1 Section 6.2 Rules of Exponents Rules of Exponents California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll multiply and divide algebraic expressions using the rules of exponents. Key words: exponent Donβt forget: See Topic 1.3.1 for more on the rules of exponents. Donβt forget: Rules 3 and 4 mean that here you multiply each power from inside the parentheses by 2, the power outside the parentheses. You learned about the rules of exponents in Topic 1.3.1 β in this Topic, youβll apply those same rules to monomials and polynomials. Use the Rules of Exponents to Simplify Expressions These are the same rules you learned in Chapter 1, but this time youβll use them to simplify algebraic expressions: Rules of Exponents 1) xaΒ·xb = xa+b 3) (xa)b = xab 2) xa Γ· xb = xa |
βb (if x Ο 0) 4) (cx)b = cbxb a b 5) x = b a x or ( b a ) x 6) xβa = 1 x a (if x Ο 0) 7) x0 = 1 Example 1 Simplify the expression (β2x2m)(β3x3m3). Solution (β2x2m)(β3x3m3) = (β2)(β3)(x2)(x3)(m)(m3) = 6x2+3Β·m1+3 = 6x5m4 Put all like variables together Use Rule 1 and add the powers Example 2 Simplify the expression (3a2xb3)2. Solution (3a2xb3)2 = 32Β·a2Β·2Β·x2Β·b3Β·2 = 9a4x2b6 Use Rules 3 and 4 274 Section 6.2 β Multiplying Polynomials Example 3 Simplify the expression β 10 β mv. Solution β 5 3 2 10 x m v β 4 2 5 x mv β 5 10 Separate the expression into parts that have only one variable Use Rule 2 and subtract the powers = 2xm2 From Rule 5, anything to the power 0 is just 1 Guided Practice Simplify each expression. 1. β3at(4a2t3) 3. (β2x2y3)3 5. (β3x2t)3(β2x3t2)2 7. 10 12 10 j k m 4 9. 2. (β5x3yt2)(β2x2y3t) 4. β2mx(3m2x β 4m2x + m3x3) 6. β2mc(β3m2c3 + 5mc) 8. 2 4 8 a b c 14 7 4 a b 4 10. 9 4 4 16 b a j 5 3 ( b a c 32 ) 2 Independent Practice Simplify. β 1. 1 βββ β 2 3 4 a b β 0 βββ β β 2. 2 βββ β 8 4 3 xy z β β 2 βββ β 3. 4a2(a2 |
β b2) 4. 4m2x2(x2 + x + 1) 5. a(a + 4) + 4(a + 4) 6. 2a(a β 4) β 3(a β 4) 7. m2n3(mx2 + 3nx + 2) β 4m2n3 8. 4m2n2(m3n8 + 4) β 3m3n10(m2 + 2n3. b a 6 19 b a Find the value of? that makes these statements true. 10. 11. 10 17 2 7 12. 4 4 20 3 a m 7 a m 6 13. m?(m4 + 2m3) = m6 + 2m5 14. m4a6(3m?a8 + 4m2a?) = 3m7a14 + 4m6a9 15 16. 4 18? 8 7 x y c? 10 x y c 6 = 2 9 yc 3 x Round Up Round Up You can apply the rules of exponents to any algebraic values. In this Topic you just dealt with monomials, but the rules work with expressions with more than one term too. Section 6.2 β Multiplying Polynomials 275 Topic 6.2.2 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll learn how to multiply monomials and polynomials. Key words: polynomial monomial distributive property degree Polynomial Multiplication Polynomial Multiplication To multiply two polynomials together, you have to multiply every single term together, one by one. The Distributive Property and Polynomial Products In Topic 6.1.2 you saw the method for multiplying a polynomial by a number β you multiply each term separately by that number. This methodβs based on the distributive property from Topic 1.2.7. In the same sort of way, when you multiply a polynomial by a monomial, you multiply each term separately by that monomial β again, |
using the distributive property. Example 1 Simplify the expression β2a(a + 3a2). Solution β2a(a + 3a2) is a product of the monomial β2a and the binomial (a + 3a2), so multiply each term of the binomial by the monomial. = β2a(a) + (β2a)(3a2) = β2a2 β 6a3 To find the product of two polynomials, such as (a β 2b)(3a + b), you use the distributive property twice: Example 2 Simplify the expression (a β 2b)(3a + b). Solution (a β 2b)(3a + b) = (a)(3a + b) + (β2b)(3a + b) = [(a)(3a) + (a)(b)] + [(β2b)(3a) + (β2b)(b)] = (3a2 + ab) + (β6ab β 2b2) = 3a2 + ab β 6ab β 2b2 = 3a2 β 5ab β 2b2 Use the distributive property twice 276 Section 6.2 β Multiplying Polynomials Example 3 Simplify (3x β 2m)(4x β 3m). Solution (3x β 2m)(4x β 3m) = 3x(4x β 3m) β 2m(4x β 3m) = 12x2 β 9mx β 8mx + 6m2 = 12x2 β 17mx + 6m2 Example 4 Simplify (v + 3)(4 + v). Solution (v + 3)(4 + v) = v(4 + v) + 3(4 + v) = 4v + v2 + 12 + 3v = v2 + 7v + 12 Guided Practice Expand and simplify each product, using the distributive method. Show all your work. 1. (m + c)(m + 2c) 3. (2x β 3)(2x + 5) 5. (3x β 5)(2x β 3) 2. (x β 3y)(x + 2y) 4. (a β 4b)(a + 3b) 6. (5x + 3y)(2x + 3y) Determine whether the following are correct for the products given. 7. (a |
+ b)(a β b) = a2 β b2 9. (a β b)2 = a2 β 2ab + b2 8. (a + b)2 = a2 + b2 10. (a + b)(a + b) = a2 + 2ab + b2 You Can Multiply Polynomials with Lots of Terms Check it out: Multiply each term in one set of parentheses by every term in the second set of parentheses. Example 5 Simplify (x + 2)(x2 + 2x + 3). Solution (x + 2)(x2 + 2x + 3) = x(x2 + 2x + 3) + 2(x2 + 2x + 3) = x3 + 2x2 + 3x + 2x2 + 4x + 6 = x3 + 4x2 + 7x + 6 Section 6.2 β Multiplying Polynomials 277 The Highest Power Gives the Degree of a Polynomial Example 6 Simplify (x β 3)(2x2 β 3x + 2) and state the degree of the product. Solution (x β 3)(2x2 β 3x + 2) = x(2x2 β 3x + 2) β 3(2x2 β 3x + 2) = 2x3β 3x2 + 2x β 6x2 + 9x β 6 = 2x3 β 9x2 + 11x β 6 The term 2x3 has the highest power, so the degree is 3. Guided Practice Expand and simplify each product, and state the degree of the resulting polynomial. 11. (x + 3)(2x2 β 3x + 1) 13. (x2 β 3x + 4)(2x + 1) 15. (3x + 4)(β2x2 + x β 2) 12. (2y β 3)(β3y2 β y + 1) 14. (3y3 + 4y β 2)(4y β 1) 16. (2x β 3)2 Determine whether the following are correct for the products given. 17. (a2 + b2)(a β ab + b) = a3 + b3 18. (a + b)(a2 β ab + b2) = a3 + b3 19. (a β b)(a2 + ab + b2) = a3 β b3 20. |
(a2 β b2)(a + ab + b) = a3 β b3 You Can Also Use the Stacking Method You can find the product of 63 and 27 by βstackingβ the two numbers and doing long multiplication: Donβt forget: The units, tens, hundreds, and thousands are in separate columns Γ 63 2 Γ 63 You can use the same idea to find the products of polynomials β just make sure you keep like terms in the same columns. Example 7 Expand and simplify the product (2x + 3y)(x + 5y). Donβt forget: Keep like terms in the same column. Solution 2x + 3yy Γ x + 5yy 10xy + 15y2 + 2x2 + 3xy + 15y2 2x2 + 13xy + 15y2 5y(2x + 3y) x(2x + 3y) 278 Section 6.2 β Multiplying Polynomials Example 8 Simplify (x β 2)(2x2 β 3x + 4). Solution 2x2 β 3x + 4 Γ x β 2 β4x2 + 6x β 8 2x3 β 3x2 + 4x βn8 2x3 β 7x2 + 10x β 8 Guided Practice β2(2x2 β 3x + 4) x(2x2 β 3x + 4) Use the stacking method to multiply these polynomials: 21. (3x + y)(x + 2y) 23. (3x2 + 2x + 3)(3x β 4) 25. (a + b)2 27. (a β b)(a + b) 29. (a + b)(a2 β ab + b2) 22. (4x + 5y)(2x + 3y) 24. (4x2 β 5x + 6)(4x + 5) 26. (a β b)2 28. (a β b)(a2 + ab + b2) 30. (a2 β b2)(a2 + b2) Independent Practice Expand and simplify each product, using the distributive method. Show all your work. 1. (2x + 8)(x β 4) 3. (x β 3)(2 β x) 5. (3x β 8)(x2 β 4x + 2) 2. (x2 + 3)(x β 2) 4 |
. (2x + 7)(3x + 5) 6. (2x β 4y)(3x β 3y + 4) Use the stack method to multiply. Show all your work. 7. (x2 β 4)(x + 3) 9. (4x2 β 5x)(1 + 2x β 3x2) 8. (x β y)(3x2 + xy + y2) 10. (x + 4)(3x2 β 2x + 5) Use these formulas to find each of the products in Exercises 11β16. (a + b)2 = a2 + 2ab + b2 (a β b)2 = a2 β 2ab + b2 (a + b)(a β b) = a2 β b2 11. (x + 2)2 13. (2x β 3)2 15. (5x + 3c)(5x β 3c) 12. (3x β 1)(3x + 1) 14. (4x + y)2 16. (8c + 3)2 Round Up Round Up Youβve seen the distributive property lots of times already, but itβs easy to make calculation errors when youβre multiplying long polynomials β so be careful and check your work. Section 6.2 β Multiplying Polynomials 279 Topic 6.2.3 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll multiply polynomials to solve problems involving area and volume. Key words: polynomial monomial distributive property Polynomial Multiplication Polynomial Multiplication β Area and Volume β Area and Volume Polynomial multiplication isnβt just about abstract math problems. Like everything in math, you can use it to work out problems dealing with everyday life. Find Areas by Multiplying Polynomials x x x x Example 1 Find the area of the space between the two rectangles: x (3 + 2) inches x x (5 + 6) inches x x Solution The length |
of the middle rectangle is 5x + 6 β 2x = (3x + 6) inches. The width of the middle rectangle is 3x + 2 β 2x = (x + 2) inches. x Area of space = area of large rectangle β area of small rectangle = (5x + 6)(3x + 2) β (3x + 6)(x + 2) = 15x2 + 10x + 18x + 12 β (3x2 + 6x + 6x + 12) = 15x2 + 28x + 12 β 3x2 β 12x β 12 = (12x2 + 16x) in2 Guided Practice 1. Find the area of a rectangle whose dimensions are (3x + 4) inches by (2x + 1) inches. 2. Find the area of the triangle shown on the right. 1 (The formula for the area of a triangle is Area = 2 bh.) 3. The height of a triangle is (3x β 2) inches and its base is (4x + 10) inches. Find the area of the triangle. 4. Find the area of a rectangle whose dimensions are (3 + 2x) inches by (5 + 6x) inches. 2x β 3 5x + 3 5. Find the area of a square with side length (a2 + b2 β c2) feet. 6. The area of a trapezoid is given by A = Β½h(b1 + b2) where h is the height and b1 and b2 are the lengths of the parallel sides. Find the area of this trapezoid. x in. (x + 1) in. (x + 4) in. 280 Section 6.2 β Multiplying Polynomials Multiply Polynomials to Find Volumes Example 2 Find the volume of the box on the right. (4 + 6) inches x (6 β 4) inches x Solution Volume = Length Γ Width Γ Height (5 + 8) inches x = (5x + 8)(6x β 4)(4x + 6) = [5x(6x β 4) + 8(6x β 4)](4x + 6) Multiply the first two polynomials = (30x2 β 20x + 48x β 32)(4x + 6) = (30x2 + 28x β 32)(4x + 6) Simplify the first product = 4 |
x(30x2 + 28x β 32) + 6(30x2 + 28x β 32) Multiply out again = 120x3 + 112x2 β 128x + 180x2 + 168x β 192 = 120x3 + 112x2 + 180x2 β 128x + 168x β 192 Commutative law = (120x3 + 292x2 + 40x β 192) in3 Example 3 Find the volume of a box made from the sheet below by removing the four corners and folding. 2x 2x 2x 2x 2x 2x 2x 2x n i 6 2x in 8 in Solution Volume = Length Γ Width Γ Height (8 β 4x) in (6 β 4x) in = (8 β 4x)(6 β 4x)(2x) = [8(6 β 4x) β 4x(6 β 4x)](2x) = (48 β 32x β 24x + 16x2)(2x) = (48 β 56x + 16x2)2x = 96x β 112x2 + 32x3 = (32x3 β 112x2 + 96x) in3 Multiply the first two polynomials Simplify the first product Multiply by the third polynomial Section 6.2 β Multiplying Polynomials 281 Guided Practice 7. Find the volume of a cube with side length (3x + 6) inches. 8. A concrete walkway around a swimming pool is 6 feet wide. If the length of the pool is twice the width, x feet, what is the combined area of the walkway and pool? 6 ft 6 ft x ft 6 ft 6 ft 2 ftx Use the rectangular prism shown to answer these questions. 9. Find the volume of the prism. 10. Find the surface area of the prism. 11. If the height of the prism was reduced by 10%, what would be the new volume of the prism? x ( + 7) ft (3 β 1) ft x (2 + 3) ft x Independent Practice Expand and simplify the following. 1. (3y + 5)3 2. (2y β 1)3 3. The area of a parallelogram is given by the formula A = bh, where b is the length of the base and h is the height of the parallelogram. Find the area of a parallelogram that has a base length of |
(2x2 + 3x β 1) cm and a height of (3x β 1) cm. 4. A gardener wants to put a walkway around her garden, as shown on the right. What is the area of the walkway? x x x x x (2 + 5) feet (7 + 3) feet x 5. Obike made a box from a 10 inch by 9 inch piece of cardboard by cutting squares of x units from each of the four corners. Find the volume of his box. x x x x 10 inches 6. Find the volume of the solid shown. 7. Find the volume of another triangular prism that has the same base measurements as this one but a height 25% less than the height shown here. x ( + 4) ft ) ft x ( + 6) ft Round Up Round Up For problems involving area, youβll have to multiply two terms or polynomials together. For problems involving volume, youβll have to multiply three terms or polynomials. 282 Section 6.2 β Multiplying Polynomials TTTTTopicopicopicopicopic 6.3.16.3.1 6.3.16.3.1 6.3.1 Section 6.3 y Monomials y Monomials vision b vision b DiDiDiDiDivision b y Monomials vision by Monomials y Monomials vision b DiDiDiDiDivision b y Monomials y Monomials vision b vision b vision by Monomials y Monomials vision b y Monomials California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll learn how to use the rules of exponents to divide a polynomial by a monomial. Key words: polynomial monomial exponent distributive property Donβt forget: You covered the rules of exponents in Topics 1.3.1 and 6.2.1. Those rules of exponents that you saw in |
Topic 6.2.1 really are useful. In this Topic youβll use them to divide polynomials by monomials. y a Monomial y a Monomial ynomial b viding a Polololololynomial b ynomial b viding a P DiDiDiDiDividing a P viding a P y a Monomial ynomial by a Monomial y a Monomial ynomial b viding a P To divide a polynomial by a monomial, you need to use the rules of exponents. The particular rule thatβs useful here is: a b = xaβb provided x Ο 0 x x Example Example Example Example Example 11111 Divide 2xΒ² by x. Solution Solution Solution Solution Solution 2 2x x = 2x2β1 = 2x1 = 2x Example Example Example Example Example 22222 Divide 2xΒ³y + xyΒ² by xy. Donβt forget: Like with multiplying, you need to treat the powers of x and y separately. Solution Solution Solution Solution Solution 2 3 x y + 2 xy xy = = β ββββ β β ββββ β 2 3 x y xy β ββββ β + 2 3 x x β
y y β ββββ β 2 β ββββ β β ββββ β xy xy β x ββββ + β
β x β ββββ β 2 y y DiDiDiDiDivide eac vide eac vide eac h ter h ter m in the m in the vide each ter h term in the m in the vide eac h ter m in the ession by y y y y xy,,,,, using ession b ession b eeeeexprxprxprxprxpression b using using using ession b using opertytytytyty oper oper e pre proper e pre pr the distributiuti |
utiutiutivvvvve pr the distrib the distrib oper the distrib the distrib = (2x3β1 β y1β1) + (x1β1 β y2β1) = (2x2 β 1) + (1 β y1) = 2x2 + y Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 283283283283283 Example Example Example Example Example 33333 Simplify mc v 2 10 4 3 mc v. Solution Solution Solution Solution Solution 2 3 2 m c v β 10 4 3 mc mc v 2 = β ββββ β 3 2 m c v 2 β 2 mc v 2 β ββββ β β β ββββ β β 2 3 m c v 4 β 2 mc v 2 β ββββ β + β ββββ β β 10 2β mc v33 4 2 ββββmc v β = (β1βm3β1) β (β2βm2β1βc3β2) + (β5βc4β2βv3β1) = βm2 + 2mc β 5c2v2 Guided Practice Simplify each of these quotients. Donβt forget: You can cancel anything that is on both the top and the bottom of the fraction. See Topic 6.3.3. 1. 9m3c2v4 Γ· (β3m2cv3) β 10 3. 3 2 m x β 5 + 15 12 5 5. β14. 4. 6. β 12 β 12 df k 3 β 12 dk + 28 8 8 ca d k 4 Independent Practice Simplify each of these quotients β 12 1. 3. 12 x 2 β 15 β 16 2 2 2 m c v 4 4. 2 3 xy xyz 5. Divide 15x5 β 10x3 + 25x2 by β5x2. 6. Divide 20a6b4 β |
14a7b5 + 10a3b7 by 2a3b4. 7. Divide 4m5x7v6 β 12m4c2x8v4 + 16a3m6c2x9v7 by β4m4x7v4. Find the missing exponent in the quotients. β 2 x y? 9. 8 xy 4x ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This leads on to the next few Topics, where youβll divide one polynomial by another polynomial. First, youβll learn how to find the multiplicative inverse of a polynomial in Topic 6.3.2. 284284284284284 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 TTTTTopicopicopicopicopic 6.3.26.3.2 6.3.26.3.2 6.3.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding finding finding ocal, taking a root, ocal, ocal, ecipr ecipr the r the r eciprocal, the recipr the r ocal, ecipr the r and raising to a fractional power..... TTTTThehehehehey under stand stand y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of eeeeexponents xponents..... xponents xponents xponents Students Students 10.0: 10.0: Students add, subtract, 10.0: Students Students 10.0: 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students |
solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll learn how to find the multiplicative inverse of a polynomial, and how to use negative exponents. Key words: polynomial monomial reciprocal exponent ynomials and ynomials and PPPPPolololololynomials and ynomials and ynomials and ynomials and ynomials and PPPPPolololololynomials and ynomials and ynomials and NeNeNeNeNegggggaaaaatititititivvvvve Pe Pe Pe Pe Pooooowwwwwererererersssss NeNeNeNeNegggggaaaaatititititivvvvve Pe Pe Pe Pe Pooooowwwwwererererersssss Before you divide one polynomial by another polynomial, you need to know how to write the multiplicative inverse of a polynomial. ocal of a P a P a P a P a Polololololynomial ynomial ynomial ocal of ocal of ecipr ecipr Finding the R Finding the R ynomial eciprocal of Finding the Recipr ynomial ocal of ecipr Finding the R Finding the R The reciprocal of a polynomial is its multiplicative inverse. To find the reciprocal of Polynomial A, you divide the number 1 by Polynomial A. Example Example Example Example Example 11111 Find the reciprocal of each of these polynomials: b) β5xΒ² + 3x β 1 a) 2x β 1 Solution Solution Solution Solution Solution a) 1 1x β 2 b) 2β 5 x 1 + β 3 x 1 By the definition of division, to divide by a polynomial you need to multiply by the reciprocal (the inverse under multiplication) of that polynomial. Guided Practice Find the multiplicative inverse of each of these expressions. 1. ab 2. a2 3. 2x + 4b 4. 3x + 1 5. 8x3 β 16x2 + 4 7. 2x2y4 8. 2 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 285285285285285 ocals as Negggggaaaaatititititivvvvve |
Exponents e Exponents e Exponents ocals as Ne ocals as Ne ecipr ecipr rite R ou Can WWWWWrite R rite R ou Can YYYYYou Can ou Can e Exponents eciprocals as Ne rite Recipr e Exponents ocals as Ne ecipr rite R ou Can This rule of exponents gives you another way to express fractions: xβa = 1 x a provided x Ο 0 Example Example Example Example Example 22222 Simplify the following expressions: a) 2β3 b) 2x β (3y)β1 c) 6b3 β (3ab)β2 b) 2x β (3y)β Solution Solution Solution Solution Solution a) 2β) 6b3 β (3ab)β2 = 3 β
6 b 1 ab Guided Practice Simplify the following expressions. 10. (2a)β2 12. (4x2y)β2 14. (2x3y)β2(3x)3 16. (15a3b4x4)(3a2b2x3)β1 18. (2a2b3c2d3)2(β 1 3 ab2cd2)β2 20. (2abc)3(3a2bc)β2(a2b3c4) 11. (3ab)β3 13. (4xy)β1(2x2) 15. (16x5y7z12)(4x3y2z8)β1 17. (40ax2y4z5)2(4xy2z4)β2 19. (3xyz)2(2xyz)(6xyz)β1 21. (4xy)(8x2y3)(x2y5)β1 286286286286286 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 ynomials as Frrrrractions actions actions ynomials as F rite Polololololynomials as F ynomials as F rite P ou Can WWWWWrite P rite P ou Can YYYYYou Can ou Can actions actions ynomials as F rite P ou Can A polynomial raised to a negative power can also be written as a fraction. Check it |
out If a pair of parentheses is raised to a negative power, its entire contents go to the bottom of the fraction. Example Example Example Example Example 33333 Simplify the following expressions: a) (a + b)β(b β c)β1 b) (m β c)(m + c)β(m β c)β1 c) (x β 1)β1β(x + 1)β1 Solution Solution Solution Solution Solution a) (a + b)β(b β c)β) (m β c)(m + c)β(m β c)β1 = ( β + m c m c )( β m c ) ( ) = m + c Check it out Thereβs more on canceling fractions in 6.3.3. c) (x β 1)β1β(x + 1)β( 1 ) Guided Practice Simplify the following expressions. 23. (w + v) Γ (w β v)β1 25. (a β x)β2(a β x) 22. (x + y) Γ (x + y)β1 24. (a β b)β1(a β b)(a + b) 26. (a β b)(a β b)(a + b)β1(a β b)β1 27. (a + x)(a β x)(a β b)β1 28. (t β s)2(t β s)β1(t + s )( 4 ) 29. (7y + 6z)(7y β 5z)β1(7y + 6z)β2 k j 4 )( β 41 ( + β 31. 30 Independent Practice Find the multiplicative inverse of the following. 1. 1.2 + 3.3a 4. 8m2 β 5mz + 4 2. 3x2 + 5y3 17 a β. 3 9 5. Simplify the following expressions. 3. β18.6 β 1.5x x. 4 3 y + 6.. 8 9 4 7. 15a2(β3a2)β1 9. (2y + 3)β2 Γ (2y + 3) 11. (2a2b4)β1(4ab)β1(16a3b4) 13. 2x(x β 1)(x β 1) β1( |
2) β1 15. (z β 5)2(3)β2(15)(z β 5)β1 8. (β2x)2(6ax)β2 10. 4ab Γ (3a)β1 Γ (4b)β1 12. (a2bβ2)3(aβ1b0)4(aβ2bβ3)β2 14. (3z)β1(z β 2)β2(z2)3(z β 2) 16. 20(z β 3)β2(z + 3)(z β 3)3(6)β1 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Now youβve got all the tools you need. Youβll start dividing one polynomial by another polynomial using two different methods in Topics 6.3.3 and 6.3.4. Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 287287287287287 TTTTTopicopicopicopicopic 6.3.36.3.3 6.3.36.3.3 6.3.3 California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll divide one polynomial by another polynomial by factoring. Key words: polynomial monomial factor exponent ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials vision b ynomials actoring actoring β Fβ Fβ Fβ Fβ Factoring actoring act |
oring β Fβ Fβ Fβ Fβ Factoring actoring actoring actoring actoring Now youβre ready to divide a polynomial by another polynomial. The simplest way to do this is by factoring the numerator and the denominator. essions essions actions Helps to Simplify Expr Canceling Frrrrractions Helps to Simplify Expr actions Helps to Simplify Expr Canceling F Canceling F essions actions Helps to Simplify Expressions essions actions Helps to Simplify Expr Canceling F Canceling F If a numerical fraction has a common factor in the numerator and denominator, you can cancel it β for example, = 5 10 β
In the same way, if there are common factors in the numerator and denominator of an algebraic fraction, you can cancel them. Example Example Example Example Example 11111 Simplify ( x + 1 2 )( + x ( + x 1 ) 3 ) Solution Solution Solution Solution Solution + ( x 1 2 )( + x ( 3 ) + x 1 ) = 2x + 3 This techniqueβs really useful for dividing polynomials when the polynomials have already been factored: Check it out: See Chapter 8 for more about canceling fractions. Example Example Example Example Example 22222 Divide (x + 4)(1 β x)(3x + 2) by (1 β x). Solution Solution Solution Solution Solution (x + 4)(1 β x)(3x + 2) Γ· (1 β xx + 4)(3x + 2) 288288288288288 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 Guided Practice Simplify each expression. + ( x ( 9 )( β x x 4 β 4 ) ) 1. + x 2 ( + )( 1 3 2. ( 2 x 1 2 )( + x β x 1 ) x 4 2 )( β 1 ) ( ( 5 x + x + )( 2 7 2 )( x x + + )( 7 2 3 5 )( x x + + 3 ) 2 ) 3. 4. ( β 3 2 + 3 x )( β. 6. ( 2 x + 7 x )( 2 β + 25 )( )( + 1 x ) 7. + 1 )( z 2 10 z β 1 ) ( Γ· 9. Divide (x |
+ 3)(x + 4) by ( x + 3 ) 8. 10. Divide 3 x x + by actorededededed actor actor ynomial Can Be F the Polololololynomial Can Be F ynomial Can Be F the P vides Evenlenlenlenlenlyyyyy,,,,, the P the P vides Ev vides Ev It Di IfIfIfIfIf It Di It Di ynomial Can Be Factor It Divides Ev actor ynomial Can Be F the P vides Ev It Di If you can divide a polynomial evenly, that means there is no remainder. This means that it must be possible to factor the polynomial (and it means that the divisor is a factor of the polynomial). Example Example Example Example Example 33333 Given that (x + 1) divides evenly into (xΒ² β 4x β 5), find (xΒ² β 4x β 5) Γ· (x + 1). Solution Solution Solution Solution Solution (xΒ² β 4x β 5) Γ· (x + 1) Check it out: In Example 3, the numerator, x2 β 4x β 5, has been factored, giving (x + 1)(x β 5). See Section 6.6 for more on factoring quadratics )( x β 5) ou know thaw thaw thaw thaw that (t (t (t (t (x + 1) is a f actor actor ou kno ou kno YYYYYou kno + 1) is a f + 1) is a f actor + 1) is a factor actor ou kno + 1) is a f vides evvvvvenlenlenlenlenlyyyyy vides e vides e e told it di e told it di because youβouβouβouβouβrrrrre told it di because y because y e told it divides e vides e e told it di because y because y ) 5 Cancel (x + 1) fr Cancel ( Cancel ( om the top and bottom om the top and bottom + 1) fr + 1) fr om the top and bottom + 1) from the top and bottom Cancel ( Cancel ( om the top and bottom + 1) fr Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 289289289289289 Check it |
out: See Sections 6.6 and 6.8 for more on factoring quadratics. Guided Practice Simplify the quotients by canceling factors. 11. + x 4 + x 8 2 2 2x 14. x + 12 x 2 x 17. 2 + β x +( 2 x 20 ) 5 12. 2x 12 x + 4 16 15. x 9 xβ 12 2 3 x 18 13. 20 2 z + + 4 5 z z 16. x 5 2 x β β 15 9 19 20. Find the ratio of the surface area to the volume of a cube with side length b. 21. Divide 4x β 12 by x2 β 2x β 3. Independent Practice Name the two factors that would divide into each expression below. 1. x2 + 7x 4. 6x2 + 9x 2. 2x β 8 5. x2 + 8x + 15 3. 6a β 15 6. a2 β 81 Simplify the quotients by canceling factors. 7. x 2 x β 2 β 4 10. a 2 a β β 8 64 8. β x 2 β x 6 5 15 4 11. β + 2a a 16 2 9. a 6 2 a 10 + + 42 70 a 12. 3 x x 2 + + x 1 13 14 15 12 6 16 12 2 x 17. 15 β β x 2 β 2 25 x 19. Find the ratio of the surface area to the volume of the rectangular prism shown. 2 + β x β 2 x 8 18. 2 x 2 b 2b b ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This methodβs most useful for βdivides evenlyβ questions β if a question mentions remainders, the long division method in Topic 6.3.4 is probably better. There are two methods for polynomial division, and you should use the one that makes the most sense for the question youβre doing. 290290290290290 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 TTTTTopicopicopicopicopic 6.3.46.3.4 6.3.46.3.4 6.3.4 California Standards: Students Students 10.0: 10.0: 10.0: Students Students |
add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: Youβll divide one polynomial by another polynomial by long division. Key words: polynomial monomial divisor dividend Donβt forget: The dididididividend vidend vidend vidend is the vidend expression that is being divided. The dididididivisor visor visor visor is the expression visor you are dividing by. ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision vision β Long Di β Long Di vision β Long Division vision β Long Di β Long Di vision vision β Long Di β Long Di β Long Division vision β Long Di vision β Long Di When a polynomial canβt be factored, there has to be some other way to divide it by another polynomial. The next method to try is the long division method. visions visions xact Di xact Di or None or None vision Method β f vision Method β f Long Di Long Di visions xact Divisions or Nonexact Di vision Method β for None Long Division Method β f visions xact Di or None vision Method β f Long Di Long Di The long division method for dividing polynomials is really similar to the long division method for integers. The aim is to find out how many groups of the divisor can be subtracted from the dividend. Example Example Example Example Example 11111 Calculate 6 2x β β 2 11 x β x 5 11. Solution Solution Solution Solution Solution First, consider the leading term of the divisor and the leading term of the dividend. 6xΒ² divided by 2x is 3x β in other words 2x will go into 6xΒ² 3x times. So 3x goes |
above the line. Leading term of dividend Leading term of divisor 3x 6 Β² β 11 β 11 x x 2 β 5x Then subtract the product of 3x and (2x β 5) from the dividend. 3x groups of (2x β 5) is (6xΒ² β 15x). Subtract 3 x groups of (2 β 5) x from the dividend, leaving a remainder of 4 β 11. x 2 β 5x 3x 6 Β² β 11 β 11 x x x β (6 Β² β 15 ) x 4 β 11 x Next see how many times the leading term of the divisor will go into the leading term of what is left of the dividend. 4x Γ· 2x = 2, so you now need to subtract 2 groups of (2x β 5). 2 β 5x 3 + 2 x 6 Β² β 11 β 11 x x x β (6 Β² β 15 ) x 4 β 11 x β (4 β 10) x β1 2 groups of (2 β 5) = x(4 β 10). x Subtracting that from what is left of the dividend leaves a remainder of β1. 6 2x So β β 2 x 11 β x 5 11 = (3x + 2) remainder β1 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 291291291291291 Guided Practice Divide using the long division method. 1. 4. 7. 2 2x 5 + β 9 x + 5 x + 16 2a 3 + 16 + a a 4 2 x x β + 25 5 2. 5. 7 2a + a + 6 a 23 + 2x 8. β 24 β x x 4 + 48 9. Divide (a3 β 6a β 4) by (a + 2). 10. Divide (4b2 + 22b + 12) by (2b + 1). 4 b 3. 6. 5 β β 2x 15 ynomials TTTTToooooooooo ynomials actor Higher-De-De-De-De-Degggggrrrrree Pee Pee Pee Pee Polololololynomials ynomials actor Higher actor Higher ou Can F YYYYYou Can F ou Can F ou Can Factor Higher ynomials actor Higher ou Can F If youβve got polyn |
omials of degree higher than 2 (such as cubic equations), itβs not always clear how to factor them. You can use long division to help factor expressions. For example, if you divide Polynomial A by Polynomial B and get an answer with remainder zero, then Polynomial B is a factor of Polynomial A. Example Example Example Example Example 22222 Divide (9mΒ³ β 3mΒ² β 26m β 8) by (3m + 4). Solution Solution Solution Solution Solution 3 + 4m β 26 β 8 m m m β (9m m β (β15m β15 Β² β 26 β 8 Β³ + 12 Β²) m Β² β 20 ) m β6 β 8m m β(β6 β 8) 0 Check it out: Factoring a polynomial fully can be useful for drawing graphs or solving equations. So (9mΒ³ β 3mΒ² β 26m β 8) Γ· (3m + 4) = 3m2 β 5m β 2 This also means that (9mΒ³ β 3mΒ² β 26m β 8) = (3m + 4)(3m2 β 5m β 2). If (3m2 β 5m β 2) can also be factored, then you can fully factor (9mΒ³ β 3mΒ² β 26m β 8). 292292292292292 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 Guided Practice Simplify each quotient by dividing using the long division method. 15 11. + 2x 3 x β 4 17 x + 4 12 13. 3 β 6 y 2 y 19 y 2 + β 14 5 β 10 y 14 15. Divide 15a2 + 7ab β 2b2 by 5a β b. 16. Divide 2x4 β 2x3 + 3x β 1 by 2x3 + 1. Find the remaining factors of the polynomial, given that: 17. (x + 3) is a factor of x2 β x β 12. 18. (x + 4) is a factor of x3 + 4x2 β 25x β 100. Independent Practice 1. Divide (x2 + 16x + 49) by (x + 4). 2. Divide (x2 + 3x β 6) by (x + 4). Carry out the following divisions |
using the long division method. 3 2 x 3. β β 2 x β x 13 3 β 6 x 4 2a 4. 2x 2a 7. + 10 + a β 1 a 1 β 6 8. 2 β 2x 3 x x 11 β 2 + 9 9. 2 x 15 + β x 2 + 5 x 10. 2a 3 1 β β 4 a + 1 a 11 2a 12. + β a 35 β a 3 4 36 13. 10 2x + 2 x + 10 x 21 + 3 14. 3 a β β 7 22 a a β β 33 a 7 15. The width of a rectangle is (x β 3) cm and the area is (3x2 β 10x + 3) cm2. What is the length? 16. The base of a triangle is (x + 2) meters and the area is (x3 β 6x β 4) meters squared. What is the height of the triangle? (Area of a triangle is A = Β½bh.) Find the remaining factors of the polynomial, given that: 17. (y + 5) is a factor of y3 + 9y2 + 23y + 15. 18. (y β 2) is a factor of y3 β 7y2 + 16y β 12. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It can sometimes take a while to calculate using long division β but sometimes itβs the only way of working out a polynomial division. If the division looks simple you should try using factoring first. Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 293293293293293 TTTTTopicopicopicopicopic 6.3.56.3.5 6.3.56.3.5 6.3.5 California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol ultisteppppp ultiste Students solve me me me me |
multiste ultiste Students solv Students solv Students solv ultiste Students solv luding worororororddddd luding w luding w inc inc lems,,,,, inc lems lems prprprprproboboboboblems including w luding w inc lems prprprprproboboboboblems lems,,,,, b b b b by using these y using these y using these lems lems y using these lems y using these hniques..... hniques hniques tectectectectechniques hniques What it means for you: Youβll use factoring and long division to solve real-life problems involving polynomial division. Key words: polynomial monomial divisor dividend Check it out: Youβll cover factoring quadratics later, in Section 6.6. Check it out: Which method you choose really just depends on how easy you find long division compared with factoring. ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b tions tions pplica pplica β β β β β AAAAApplica tions pplications tions pplica β β β β β AAAAApplica tions tions pplica pplica pplications tions pplica tions You can use most of the skills youβve learned in this Section to solve geometric problems involving polynomials. lems Often Invvvvvolvolvolvolvolve Pe Pe Pe Pe Polololololynomials ynomials ynomials lems Often In olume Proboboboboblems Often In lems Often In olume Pr VVVVVolume Pr olume Pr ynomials ynomials lems Often In olume Pr Example Example Example Example Example 11111 The volume of the box shown is (40xΒ³ + 34xΒ² β 5x β 6) cmΒ³. Write an expression for the height in cm, h, of the box, if it has width (2x + 1) cm and length (4x + 3) cm. |
(2x Solution Solution Solution Solution Solution Start by writing out the formula for volume: + 1) c m h cm (4x + 3) cm volume = length Γ width Γ height height = volume Γ length width h = 40 3 x ( 4 + 2 34 + x )( There are a couple of different ways to tackle a division problem like this: 1) You could use long division to divide the volume by one of the factors, leaving you with a quadratic expression that you can then factor by trial and error β you already know one of the factors. or 2) You could multiply together the two factors you know to get a quadratic expression. You can then use long division to divide the volume by the quadratic expression. In this case, the long division is a bit more complicated than those we tackled on the previous page, but this method does the division in one step rather than two. 294294294294294 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 Example Example Example Example Example 22222 Method 1: Use long division to find the height, h, of the box shown in Example 1. Solution Solution Solution Solution Solution OK, so you already have the expression h = 3 40 x ( 4 2 + 34 x + )(. Use long division to divide the volume by one of the factors: 4 + 3 x β (40 Β³ + 30 Β²) x β 2 x x 10 Β² + 40 Β³ + 34 4 Β² + 3 ) β8 β 6x x β (β8 β 6) 0 Then you just need to factor the quadratic (10xΒ² + x β 2). You know that one of the factors is (2x + 1), from the original question. By trial and error, you can work out that (10xΒ² + x β 2) will factor into (2x + 1)(5x β 2). So the height of the box must be (5x β 2) cm. Alternatively, you could do another long division to divide (10xΒ² + x β 2) by (2x + 1). Example Example Example Example Example 33333 Method 2: Use the multiplying factors method to find the height, h, of the box shown in Example 1. Check it out: Again, you start with the leading term of each polynomial. Solution Solution Solution Solution Solution Start by multiplying together the factors you know |
: (4x + 3)(2x + 1) = 8xΒ² + 4x + 6x + 3 = 8xΒ² + 10x + 3 Then use long division to divide the volume by the product of the two factors: 8 Β² + 10 + 3 x x 5 β 2 40 Β³ + 34 40 Β³ + 50 Β² + 15 ) x x x β16 Β² β 20 β 6 β (β16 Β² β 20 β 6) x x x x So the answer is h = (5x β 2) cm. 0 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 295295295295295 Guided Practice 1. If the area of a rectangle is (6y2 + 29y + 35) square units, and its length is (3y + 7) units, find the width of the rectangle. 2. The volume of a rectangular box is (60x3 + 203x2 + 191x + 36) cubic inches. Find the height of the box if the area B of the base is (15x2 + 47x + 36) square inches. [Hint: V = Bh] 3. The area of a circle is (px2 β 10px + 25p) m2. Show that the radius is (x β 5) m. 4. The volume of a prism is (x3 + 8x2 + 19x + 12) m3. Find the area of the base if the height is (x + 4) m. 5. A rectangular prism has volume (2b3 β 7b2 + 2b + 3) m3, height (b β 1) m, and length (2b + 1) m. Find its width. Independent Practice 1. The width of a rectangle is (3x + 4) m. If the area of the rectangle is (6x2 + 5x β 4) m2, what is the length? 2. The area, A, of a triangle is given by the formula 2A = bh, where b is the base length and h is the height. Find the length of the base if A = (3x2 β 16x + 16) ft2 and h = (2x β 8) ft. 3. Find the height of a rectangle with area (2x2 β 9x + 9) ft2 and width (2x β 3 |
) ft. 4. The volume of a prism is (2x3 + x2 β 3x) m3. If the area of the base is (x2 β x) m2, what is the height of the prism? 5. A rectangular prism has volume (b3 + 9b2 + 26b + 24) m3, width (b + 2) m, and length (b + 4) m. Find its height. 6. The volume of a prism is (144s3 + 108s2 β 4s β 3) m3. If the area of the base is (36s2 β 1) m2, what is the height? 7. The volume of a cylinder is (2pp3 + 7pp2 + 8pp + 3p) in3. Find the area of the base if the height is (2p + 3) in. 8. The volume of a cylinder is (18py3 β 3py2 β 28py β 12p) ft3. Find the height of the cylinder if the area of the base is (9py2 + 12py + 4p) ft2. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In this Section youβve seen two good ways of dividing one polynomial by another polynomial β factoring and long division. 296296296296296 Section 6.3 Section 6.3 Section 6.3 β Dividing Polynomials Section 6.3 Section 6.3 Topic 6.4.1 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. 11.0 Students apply basic factoring techniques to second and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to use special cases of binomial multiplication to save time in calculations. Key words: binomial difference of two squares Check it out: |
(a + b)Β² expands to give a perfect square trinomial β see Topic 6.8.2. Section 6.4 Special Products of Special Products of Two Binomials Two Binomials This Topic is all about special cases of binomial multiplication. Knowing how to expand these special products will save you time when youβre dealing with binomials later in Algebra I. Remember These Three Special Binomial Products Given any real numbers a and b, then: (a + b)2 = a2 + 2ab + b2 (a β b)2 = a2 β 2ab + b2 (a + b)(a β b) = a2 β b2 When You Expand (a + b)Β², You Always Get an ab-Term (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 Using the distributive property = a2 + 2ab + b2 You can relate this equation to the area of a square: a a a2 ab ab b2 b b (a + b)2 is the same as the area of this large square β add the areas of the two smaller squares, a2 and b2, and the two rectangles, 2 Γ ab. Example 1 Expand and simplify (2x + 3)2. Solution Put the expression in the form (a + b)(a + b): (2x + 3)2 = (2x + 3)(2x + 3) = 4x2 + 6x + 6x + 9 = 4x2 + 12x + 9 2x 2x 4x2 6x (2x + 3)2 is the same as the area of the large square β add the areas of the two smaller squares, 4x2 and 9, and the two rectangles, 2 Γ 6x. 6x 3 9 3 Section 6.4 β Special Products of Binomials 297 Guided Practice Find and simplify each product. 1. (x + 5)2 4. (9y + z)2 2. (2y + 3)2 5. (n2 + 4n)2 3. (3y + 1)2 6. (3t + 6b)2 Find the areas of the squares below. 7. 8. 9. 10 Check it out: (a β b)Β² also expands to give a perfect square trinomial β see Topic 6.8.2. |
When You Expand (a β b)Β², the ab-Term is Negative (a β b)2 = (a β b)(a β b) = a2 β ab β ba + b2 Using the distributive property = a2 β 2ab + b2 You can also relate this equation to the area of a smaller square: a (a β b)2 a (a β b)2 is the same as the area of the darker square. To find the area of the darker square, you can subtract two rectangles of area ab β but then you have to add back on an area of b2 (this small square). b b2 b Example 2 Expand (3y β 2)2. Solution Put the expression in the form (a β b)(a β b): (3y β 2)2 = (3y β 2)(3y β 2) = (3y)2 β 2(3y Γ 2) + 22 = 9y2 β 2(6y) + 22 = 9y2 β 12y + 4 In the diagram on the right, (3y β 2)2 is the same as the area of the darker square. 3y (3y β 2)2 3y 2 4 2 298 Section 6.4 β Special Products of Binomials Guided Practice Find and simplify each product. 11. (m β 4)2 14. (2a β 5)2 17. (x2 β y)2 12. (y β 2)2 15. (7d β f)2 13. (4k β 5)2 16. (xy β 3)2 18. (9x β 3y)2 19. (3x2 β 4)2 20. Find the area of the square shown. (3x β 1) in. Check it out: This is called a difference of two squares, and it can also be used in factoring expressions β see Topic 6.8.1. Thereβs No ab-Term When You Expand (a + b)(a β b) (a + b)(a β b) = a2 β ab + ba β b2 Using the distributive property = a2 β ab + ab β b2 = a2 β b2 The fact that no ab-term is left at all makes it unusual, but also very useful if you remember it. Example 3 Multiply out (4m + 3)(4m β 3). Solution |
The expression is already in the form (a + b)(a β b), so you can convert it to the form a2 β b2: (4m + 3)(4m β 3) = (4m)2 β 32 = 16m2 β 9 Guided Practice Find and simplify each product. 21. (m β v)(m + v) 23. (3y + x)(3y β x) 22. (x + 5)(x β 5) 24. (k β 6t)(k + 6t) 25. (3x β 9y)(3x + 9y) 26. (6x + 6y)(6x β 6y) 27. (x + 3x2)(x β 3x2) 29. (x2 β x)(x2 + x) 28. (9p2 β 2)(9p2 + 2) 30. (7a2 + b)(7a2 β b) Section 6.4 β Special Products of Binomials 299 You Can Use These Standard Equations as Shortcuts The good thing about knowing these standard equations is that you donβt need to do all the work each time β you can save time by using the three special products. Example 4 Find the area of a square which has side lengths of (3x + 4) inches. Solution Using (a + b)2 = a2 + 2ab + b2 and putting in (3x + 4) instead of (a + b) you get: 3x 3x 9x2 12x Check it out: In this case, a2 = (3x)2, 2ab = 2 Γ (3x Γ 4), and b2 = 42. Area of square = (3x + 4)2 = 9x2 + (2 Γ 12x) + 16 = (9x2 + 24x + 16) in2 12x 16 4 4 Example 5 Multiply (4x β 7) by (4x β 7). Solution Using (a β b)2 = a2 β 2ab + b2 and putting in (4x β 7) instead of (a β b) you get: (4x β 7)2 = 16x2 β (2 Γ 28x) + 49 = 16x2 β 56x + 49 Example 6 Find the area of this rectangle: (5y + 4) cm Solution Using (a + b)(a β b) = a |
2 β b2 and putting in 5y for a and 4 for b you get: (5y β 4) cm Area = (5y + 4)(5y β 4) = (5y)2 β 42 = (25y2 β 16) cm2 Check it out: Using the equation here is a little shorter than expanding the parentheses. 300 Section 6.4 β Special Products of Binomials Independent Practice Find the areas of these shapes. 3. 4. 2a β b 3a β b 2 +a b 3a β b Find and simplify each product: 5. (2r β 3)(2r + 3) 6. (x3 + 2)(x3 β 2) 7. (6x2 β 1)2 8. (z2 β z)2 9. (x2 β 3x)(x2 + 3x) 10. (2x2 β 3x)(2x2 + 3x) 11. Find the area of this shaded region5x β 4) in. (5x + 4) in. The area of a circle with radius r is given by the formula A = pr2. Find the areas of these circles, giving your answers in terms of p: 12. A circle with radius (3x + 4). 13. A circle with radius (2x β 7). 14. A circle with radius (2a + b). 15. Find the coefficient of ab in the product (5a β 4b)2. 16. Find the coefficient of mc in the product (4m β 3c)(4m + 3c). Find and simplify each product: 17. (3x + 7)2 β (3x β 7)2 18. (2xa + 1)2 19. (km β xm)2 20. Nicole has a circle of card with a radius of (4x + 5) cm. She makes the circle into a ring by cutting a circular hole in the middle with a radius of (4x β 5) cm. Find the area of the ring, leaving your answer in terms of p. Round Up Round Up The main reason for learning these special products is to make your life easier when you multiply two binomials or factor quadratics. Youβll come across them throughout the rest of Algebra I. Section 6.4 β Special Products of Binomials 301 TTTTTopicopicopicopicopic 6.5.16.5.1 |
6.5.16.5.1 6.5.1 Section 6.5 Monomials Monomials s of s of actor actor FFFFFactor Monomials s of Monomials actors of Monomials s of actor FFFFFactor Monomials Monomials s of s of actor actor s of Monomials actors of Monomials s of Monomials actor California Standards: 11.0 Students applpplpplpplpply basic y basic y basic 11.0 Students a 11.0 Students a y basic 11.0 Students a y basic 11.0 Students a fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc tectectectectechniques inc hniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials..... What it means for you: Youβll learn how to find factors of a monomial. Key words: factor monomial greatest common factor prime factor Donβt forget: A monomial is a single term expression made up of a number or a product of a number and variables β such as 13, 2x2 or βx3yn4. In previous Topics youβve already done lots of manipulation of polynomials β but you can often make manipulations easier by breaking down polynomials into smaller chunks. In this Topic youβll break down monomials by factoring. s β and So Do |
Monomials s β and So Do Monomials actor s Has Havvvvve Fe Fe Fe Fe Factor actor s Has Ha Number Number s β and So Do Monomials actors β and So Do Monomials Numbers Ha s β and So Do Monomials actor Number Number Sometimes a number can be written as the product of two or more smaller numbers. Those smaller numbers are called factors of that number. For example, 6 = 2 Γ 3 β so 2 and 3 are factors of 6. The same is true for monomials. Unless theyβre prime numbers, monomials can be written as the product of two or more numbers or letters. Those smaller numbers or letters are called factors of that monomial. For example, 3xy = 3 Γ x Γ y β so 3, x, and y are factors of 3xy. Guided Practice Write down all of the factors of each of these numbers: 1. 8 4. 16 3. 15 6. 24 2. 10 5. 11 Write down each monomial as a product of the smallest possible factors: 7. 3x 10. 5xy 9. 6p 12. 20mn 8. 7z 11. 12uv h Monomial h Monomial Eac Eac visor of visor of he GCF is a Di TTTTThe GCF is a Di he GCF is a Di h Monomial Each Monomial visor of Eac he GCF is a Divisor of h Monomial Eac visor of he GCF is a Di The greatest common factor (GCF) of a set of monomials is the largest possible divisor of all monomials in the set. Check it out: Factors that canβt be factored themselves, like 2, 3, 5, x and y, are called prime factors. Itβs these prime factors you need to use here. Example Example Example Example Example 11111 Find the greatest common factor of 12xΒ²yΒ², 18xΒ³yΒ², and 30x4y4. Solution Solution Solution Solution Solution Start by writing down each monomial as a product of the smallest factors possible: 12 Β² Β² = 2 Γ 2 Γ 3 Γ Γ Γ Γ 18 Β³ Β² = 2 Γ 3 Γ 3 Γ Γ Γ Γ Γ 30 Then list all the numbers that are factors of all three terms: 2, 3, x, x, y, y. These are called the common factors. |
302302302302302 Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 Example 1 continued The greatest common factor is the product of all the common factors: GCF = = 6xΒ²yΒ² In other words, 6xΒ²yΒ² is the largest possible divisor of 12xΒ²yΒ², 18xΒ³yΒ², and 30x4y4. Guided Practice Use the method from Example 1 to write down the greatest common factor of each set of products: 13. 12, 24, 42 14. 9ab2, 15a2b2, 12ab 15. 6m2cv2, 10m2c2v, 4m2c2v3 16. 5mx2t, 15m2xt2, 20mxt 17. 9x3y2, 27x2y3 o or MMMMMororororore e e e e MMMMMonomials onomials onomials o or ind the GCF of TTTTTwwwwwo or o or ind the GCF of y to FFFFFind the GCF of ind the GCF of y to Another WWWWWaaaaay to y to Another Another onomials onomials o or ind the GCF of y to Another Another You can also find the GCF of two or more monomials by simply multiplying together the GCFs of each of the different parts. Example Example Example Example Example 22222 Find the greatest common factor of 12xΒ²yΒ², 18xΒ³yΒ², and 30x4y4. Solution Solution Solution Solution Solution The GCF of 12, 18, and 30 is 6. The GCF of xΒ², xΒ³, and x4 is xΒ². The GCF of yΒ² and y4 is yΒ². So, the GCF of 12xΒ²yΒ², 18xΒ³yΒ², and 30x4y4 is 6 Γ xΒ² Γ yΒ² = 6xΒ²yΒ². Check it out: You could also calculate all of these GCFs using the method in Example 1. Guided Practice Use the method from Example 2 to write down the greatest common factor of each set of products: 18. b3m2cv, bm2v 19. 2(m + 1), β3(m + 1), (m + 1)2 20. 8(v β 1)2, 4( |
v β 1)3, 12(v β 1)2 21. 6x2yz, 15xz 22. 21x4y4z4, 42x3y4z5, 14x6y3z2 Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 303303303303303 Independent Practice Write down all of the factors of each of these numbers: 1. 25 4. 67 3. 36 6. 70 2. 12 5. 80 Write each of these as a product of prime factors: 7. 48 10. 66 8. 72 11. 450 9. 120 12. 800 Check it out: You can use either method of finding the greatest common factor here β itβs up to you. Write each monomial as a product of the smallest possible factors: 13. 66z2 16. β98a2b 19. 3x2yz 15. β102x3y 18. β80rs5 21. β100f 2gh4 14. 4b3d2 17. 64y3z3 20. 16pq3r3 Write down the greatest common factor of each set of products: 22. 18, 36 24. 95, 304 26. 21p2q, 35pq2 28. β60r2s2t2, 45r3t3 30. 14a2b3, 20a3b2c2, 35ab3c2 32. 14a2b2, 18ab, 2a3b3 23. 84, 75 25. 17a, 34a2 27. 12an2, 40a4 29. 18, 30, 54 31. 18x2, 30x3y2, 54y3 33. 32m2n3, 8m2n, 56m3n2 34. The area of a rectangle is 116 square inches. What are its possible whole number dimensions? 35. The area of a rectangle is 1363 square centimeters. If the measures of the length and width are both prime numbers, what are the dimensions of the rectangle? 36. Marisela is planning to have 100 tomato plants in her garden. In what ways can she arrange them in rows so that she has the same number of plants in each row, at least 5 rows of plants, and at least 5 plants in each row? 37. A walkway is being paved using 2-ft-by-2-ft paving stones |
. If the length of the walkway is 70 ft longer than the width and its area is 6000 ft2, how many paving stones make up the length and the width of the walkway? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Factoring is the best way of working out which smaller parts make up a number or monomial. In the next few Topics youβll use factoring to break down full expressions, which makes it much easier to do tricky jobs like solving some kinds of equations. 304304304304304 Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 TTTTTopicopicopicopicopic 6.5.26.5.2 6.5.26.5.2 6.5.2 California Standards: 11.0 Students applpplpplpplpply basic y basic y basic 11.0 Students a 11.0 Students a y basic 11.0 Students a y basic 11.0 Students a fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc tectectectectechniques inc hniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial ynomial ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial ter ynomial ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to find factors of a polynomial. Key words: factor monomial polynomial |
greatest common factor actorsssss actor actor Simple F Simple F Simple Factor actor Simple F Simple F actorsssss actor actor Simple F Simple F Simple Factor actor Simple F Simple F ynomials ynomials ofofofofof P P P P Polololololynomials ynomials ynomials ofofofofof P P P P Polololololynomials ynomials ynomials ynomials ynomials The Topics in this Section are all about making your work a lot easier. Factoring polynomials can make them simpler, which means theyβre easier to manipulate. actored ed ed ed ed TTTTToooooooooo actor actor ynomials Can (Sometimes) Be F PPPPPolololololynomials Can (Sometimes) Be F ynomials Can (Sometimes) Be F ynomials Can (Sometimes) Be Factor actor ynomials Can (Sometimes) Be F When a polynomial can be expressed as the product of two or more numbers, monomials, or polynomials, those smaller numbers, monomials, or polynomials are called factors of that polynomial. For example, if Polynomial A can be written as either: 2xΒ² + x or x(2x + 1), this means that x is a factor of Polynomial A. It also means that (2x + 1) is a factor of Polynomial A. To find the factors of a polynomial, start by finding the greatest common factor of all the terms. Example Example Example Example Example 11111 Factor 6mΒ³ β 4mΒ². Solution Solution Solution Solution Solution Γ The factors that are present in both terms are 2, m, and m β so 2, m, and m are factors of both terms. So the GCF = 2 Γ m Γ m = 2mΒ² Next you need to rewrite the expression with the factor taken out: 6mΒ³ = 2 Γ m Γ m) Γ 3 Γ m = 2mΒ² Γ 3m 4mΒ² = 2 Γ 2 Γ m Γ m = (2 Γ m Γ m) Γ 2 = 2mΒ² Γ 2 So 6mΒ³ β 4mΒ² = (2mΒ² Γ 3m) β (2mΒ² Γ 2) = 2mΒ²(3m β 2) Itβs always worth checking your factoring by multiplying out again: 2mΒ²(3m β 2) |
= (2 Γ mΒ² Γ 3 Γ m) β (2 Γ mΒ² Γ 2) = (6 Γ m2+1) β (4 Γ mΒ²) = 6m3 β 4mΒ² Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 305305305305305 Guided Practice In each polynomial, find the greatest common factor of the terms. 1. 12y2 β 3y 2. a3 + 3a 3. 14a3 β 28a2 + 56a 4. 16y2 β 24y3 5. 60x3 + 24x2 + 16x 6. 18y5 + 6y4 + 3y2 Use your answers from Exercises 1 β 6 to factor the following: 7. 12y2 β 3y 8. a3 + 3a 9. 14a3 β 28a2 + 56a 10. 16y2 β 24y3 11. 60x3 + 24x2 + 16x 12. 18y5 + 6y4 + 3y2 actorsssss actor actor aking Out F e Method for or or or or TTTTTaking Out F aking Out F e Method f An An An An An AlterAlterAlterAlterAlternananananatititititivvvvve Method f e Method f aking Out Factor actor aking Out F e Method f Again, find the GCF by multiplying together the GCFs of each of the different parts β the coefficients and the variables. Then just rewrite the expression with the GCF taken out, as before. Example Example Example Example Example 22222 Factor 6mΒ³ β 4mΒ². Solution Solution Solution Solution Solution The greatest common factor of 6 and 4 is 2. Since both terms contain m, the lowest power will be a factor. So GCF = 2 Γ m Γ m = 2mΒ² To write 6mΒ³ β 4mΒ² as a factored expression, multiply and divide by 2mΒ², as shown: Check it out: Example 2 covers the same problem as Example 1, but this time using the GCF of each of the different parts. β 6mΒ³ β 4mΒ² = 2m2 6 ββββ β = 2m2(3m β 2) β ββββ β Gu |
ided Practice Factor each polynomial below. 13. x2 β 4x 15. 24x3 β 15x2 + 6x 17. 4a3 β 6a2 + 6a 19. 6b3 β 3b2 + 12b 14. x2 β x 16. 8x3 + 2x2 + 4x 18. 14b2 + 7b β 21 20. a4 + a5 + 5a3 306306306306306 Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 actors s s s s TTTTToooooooooo actor actor ynomial F e Out Polololololynomial F ynomial F e Out P ou Can TTTTTakakakakake Out P e Out P ou Can YYYYYou Can ou Can ynomial Factor actor ynomial F e Out P ou Can At the start of the Topic you saw that 2xΒ² + x can be written as x(2x + 1). This means that both x (a monomial) and 2x + 1 (a polynomial) are factors of 2xΒ² + x. The next few examples are about finding polynomial factors of the form (ax + b). Example Example Example Example Example 33333 Factor (d β 1)xΒ² + (d β 1)x + (d β 1). Solution Solution Solution Solution Solution Each term is a product of (d β 1) and something else, so (d β 1) is a common factor. To write (d β 1)xΒ² + (d β 1)x + (d β 1) as a factored expression, put the (d β 1) outside parentheses and divide everything inside the parentheses by (d β 1), as shown: (d β 1)xΒ² + (d β 1)x + (d β 1) = (d β 1) β ββββ β d β 1)(x2 + x + 1) β ββββ β Example Example Example Example Example 44444 Show that (a + b) is a factor of ac + bc + ad + bd. Solution Solution Solution Solution Solution Youβve been given the factor, so try writing the polynomial as a factored expression. If you can do that, |
youβll have shown that (a + b) is a factor. Take the (a + b) outside parentheses, as above: + ) ad + a ( b ) + bd + a ( b ) β ββββ β = + a ( ) ( ) b b a + ac + ac + bc + ad + bd β βββ β β βββ β ββ βββ β + + ac ( a bcc ) bc + ( a b + + bd ) b ad ( β ββββ β β ββββ β = (a + b)(c + d) Therefore (a + b) is a factor. Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 307307307307307 Check it out: Youβre aiming to cancel (a + b) from the top. Check it out: Note that grouping the fractions differently helps to show (a + b) as a factor. Example Example Example Example Example 55555 Factor and simplify the following expression: (x β 2)(x + 2)x + (x β 2)x + (x β 2)(x + 2). Solution Solution Solution Solution Solution Each term is a product of (x β 2) and something else, so (x β 2) is a common factor. To write (x β 2)(x + 2)x + (x β 2)x + (x β 2)(x + 2) as a factored expression, write the (x β 2) outside parentheses, then divide all terms by (x β 2): (x β 2)(x + 2)x + (x β 2)x + (x β 2)(x + 2) = (x β 2) ( x β βββ β β + x )( )( β 2 ) + 2 ) β ββββ β = (x β 2)[(x + 2)x + x + (x + 2)] = (x β 2 |
)(x2 + 2x + x + x + 2) = (x β 2)(x2 + 4x + 2) Guided Practice Factor and simplify. 21. x(2x + 1) + 3(2x + 1) 22. 3y2(2 β 3x) + y(2 β 3x) + 5(2 β 3x) 23. 2x4(5x β 3) β x2(5x β 3) + (5x β 3) 24. 2a(3a β 1) + 6(3a β 1) 25. (4 β x)x2 + (4 β x)2x + (4 β x)1 26. Show that (x + 3) is a factor of x2 + 2x + 3x + 6. 27. Show that (y + 2) is a factor of y2 + y + 2y + 2. 28. Show that (3x β 4) is a factor of 6x2 + 9x β 8x β 12. Factor and simplify the following expressions. 29. (x + 3)x + (x + 3)(x β 1) + (x + 3)x β 2(x + 3) 30. (2x β 1)2x + (2x β 1)(2x β 1) + (2x β 1)x β (2x β 1) 31. (x3 + x2)(x + 1) β (x3 + x2)(x2 + x) + (x3 + x2)(x2 + 2x + 3) 308308308308308 Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 Independent Practice In each polynomial, find the greatest common factor of the terms. 1. 8x2 β 12yx 2. 81x3 + 54x2 3. 21x3yz β 35x2y + 70xyz Factor each polynomial below. 4. 2x β 6 5. 6x2 β 12x 6. 5(c + 1) β 2y(c + 1) 7. 6y2 β 12y3 + 18y 8. x2(k + 3) + (k + 3) 9. k(y β 3) β m(y β 3) 10. (x + 1)2 β 2(x + 1)3 11. |
m(y β 5)2 β (y β 5) 12. (x2 β 2x) + (4x β 8) 13. (y2 + 3y) + (3y + 9) 14. (2my β 3mx) + (β4y + 6x) 15. β2m(x + 1) + k(x + 1) 16. x5 + 3x3 + 2x4 18. 3x3 β 6x2 + 9x 20. 2m3n β 6m2n2 + 10mn 17. 8y2x + 4yx2 + 4y2x2 19. 4x5 β 4x3 + 16x2 21. Show that (x + 4) is a factor of x2 β 5x + 4x β 20. 22. Show that (2x + 5) is a factor of 2x2 β 2x + 5x β 5. 23. Show that (3x β 1) is a factor of 6x2 + 3x β 2x β 1. 24. Show that (a β b) is a factor of 2ac β ad β 2bc + bd. Factor and simplify the following expressions. 25. (4x2 + 3)(2x + 1) + (4x2 + 3)(2x + 2) + 8(4x2 + 3) 26. 20(4a + 3b) β (x + 1)(4a + 3b) + (x2 + x + 8)(4a + 3b) + (4a + 3b) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The greatest common factor is really useful when youβre trying to factor polynomials, because itβs always the best factor to use. In the next Section youβll see that you can also factor more complicated polynomials like quadratics. Section 6.5 Section 6.5 Section 6.5 β Factors Section 6.5 Section 6.5 309309309309309 TTTTTopicopicopicopicopic 6.6.16.6.1 6.6.16.6.1 6.6.1 Section 6.6 actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFact |
oring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr actoring Quadr California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to factor simple quadratic expressions. Key words: quadratic polynomial factor binomial Check it out: The order of the parentheses doesnβt matter β you could have (x + 3) first. In Section 6.5 you worked out common factors of polynomials. Factoring quadratics follows the same rules, but you have to watch out for the squared terms. actorsssss o or More Fe Fe Fe Fe Factor actor actor o or Mor oducts of TTTTTwwwwwo or Mor o or Mor oducts of oducts of ynomials as Pr PPPPPolololololynomials as Pr ynomials as Pr ynomials as Products of actor o or Mor oducts of ynomials as Pr A quadratic polynomial has degree two, such as 2x2 β x + 7 |
or x2 + 12. Some quadratics can be factored β in other words they can be expressed as a product of two linear factors. Suppose x2 + bx + c can be written in the form (x + m)(x + n). Then: x2 + bx + c = (x + m)(x + n) b, c, m, and n are numbers = x(x + n) + m(x + n) = x2 + nx + mx + mn Expand out the parentheses using the distributive property So, x2 + bx + c = x2 + (m + n)x + mn Therefore b = m + n and c = mn So, to factor x2 + bx + c, you need to find two numbers, m and n, that multiply together to give c, and that also add together to give b. Example Example Example Example Example 11111 Factor x2 + 5x + 6. Solution Solution Solution Solution Solution The expression is x2 + 5x + 6, so find two numbers that add up to 5 and that also multiply to give 6. The numbers 2 and 3 multiply together to give 6 and add together to give 5. You can now factor the quadratic, using these two numbers: x2 + 5x + 6 = (x + 2)(x + 3) To check whether the binomial factors are correct, multiply out the parentheses and then simplify the product: (x + 2)(x + 3) = x(x + 3) + 2(x + 3) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 + 3x + 2x + 6 = x2 + 5x + 6 This is the same as the original expression, so the factors are correct. 310310310310310 Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 Check it out: The numbers β3 and +2 multiply together to give β6 and add together to give β1. Check it out: The numbers β2 and β3 multiply together to give +6 and add together to give β5. Check it out: The numbers +4 and β2 multiply together to give β8 and add together to give +2 |
. Example Example Example Example Example 22222 Factor x2 β x β 6. Solution Solution Solution Solution Solution Find two numbers that multiply to give β6 and add to give β1, the coefficient of x. Because c is negative (β6), one number must be positive and the other negative. x2 β x β 6 = (x β 3)(x + 2) Check whether the binomial factors are correct: (x β 3)(x + 2) = x(x + 2) β 3(x + 2) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 + 2x β 3x β 6 = x2 β x β 6 This is the same as the original expression, so the factors are correct. Example Example Example Example Example 33333 Factor x2 β 5x + 6. Solution Solution Solution Solution Solution Find two numbers that multiply to give +6 and add to give β5, the coefficient of x. Because c is positive (6) but b is negative, the numbers must both be negative. x2 β 5x + 6 = (x β 2)(x β 3) Check whether the binomial factors are correct: (x β 2)(x β 3) = x(x β 3) β 2(x β 3) Using the distrib Using the distrib Using the distrib Using the distributiutiutiutiutivvvvve pre pre pre pre proper oper oper opertytytytyty Using the distrib oper = x2 β 3x β 2x + 6 = x2 β 5x + 6 This is the same as the original expression, so the factors are correct. Example Example Example Example Example 44444 Factor x2 + 2x β 8. Solution Solution Solution Solution Solution Find two numbers that multiply to give β8 and add to give +2, the coefficient of x. x2 + 2x β 8 = (x + 4)(x β 2) Check whether the binomial factors are correct: (x + 4)(x β 2) = x(x β 2) + 4(x β 2) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 β 2x + 4x |
β 8 = x2 + 2x β 8 This is the same as the original expression, so the factors are correct. Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 311311311311311 Guided Practice Factor each expression below. 1. a2 + 7a + 10 3. x2 β 17x + 72 5. b2 + 2b β 24 7. x2 β 15x + 54 9. t2 + 16t + 55 11. x2 β 3x β 18 13. x2 β 2x β 15 15. x2 β 4 17. 4x2 β 64 19. x2 β 49a2 2. x2 + 7x + 12 4. x2 + x β 42 6. a2 β a β 42 8. m2 + 2m β 63 10. p2 + 9p β 10 12. p2 + p β 56 14. n2 β 5n + 4 16. x2 β 25 18. 9a2 β 36 20. 4a2 β 100b2 Independent Practice Find the value of? in the problems below. 1. x2 + 3x β 4 = (x + 4)(x +?) 3. x2 + 16x β 17 = (x +?)(x β 1) 5. a2 + 6a β 40 = (a β 4)(a +?) Factor each expression below. 2. a2 β 2a β 8 = (a +?)(a + 2) 4. x2 β 14x β 32 = (x + 2)(x +?) 6. x2 β 121 9. x2 + 2x + 1 12. d2 + 21d + 38 15. a2 β 16a + 48 18. a2 + 5a β 24 7. 4c2 β 64 10. x2 + 8x + 16 13. x2 β 13x + 42 16. x2 + 18x + 17 19. b2 β 19b β 120 8. 16a2 β 225 11. b2 β 10b + 25 14. a2 β 18a + 45 17. x2 β 24x + 80 20. x2 + 14x β 72 21. Determine whether (x + 3) is a factor of x2 β 2x β 15. 22. If 2n3 β 5 is a factor of 12n5 + 2n4 β 30 |
n2 β 5n, find the other factors. 23. If (8n β 3) is a factor of 8n3 β 3n2 β 8n + 3, find the other factors. 24. If (2x + 5) is a factor of 2x3 + 15x2 + 13x β 30, find the other factors. 25. If (a β 1) is a factor of a3 β 6a2 + 9a β 4, find the other factors. 26. If (x β 2) is factor of x3 + 5x2 β 32x + 36, find the other factors. What can you say about the signs of a and b when: 27. x2 + 9x + 16 = (x + a)(x + b), 28. x2 β 4x + 6 = (x + a)(x + b), 29. x2 + 10x β 75 = (x + a)(x + b), 30. x2 β 4x β 32 = (x + a)(x + b)? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The method in this Topic only works for quadratic expressions that have an x2 term with a coefficient of 1 (so itβs usually written just as x2 rather than 1x2). In the next Topic youβll see how to deal with other types of quadratics. 312312312312312 Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 TTTTTopicopicopicopicopic 6.6.26.6.2 6.6.26.6.2 6.6.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee |
pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to factor more complicated quadratic expressions. Key words: quadratic polynomial factor trial and error actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr + + Β² + Β² + β β β β β axΒ² + + c Β² + bx + + Β² + β β β β β axΒ² + + + Β² + Β² + Β² + bx + + c Β² + + The method in Topic 6.6.1 for factoring a quadratic expression only works if the xΒ²-term has a coefficient of 1. Itβs a little more complicated when the xΒ²-coefficient (a) isnβt 1 β but only a little. om Each h h h h TTTTTererererermmmmm om Eac om Eac actor fr actor fr e Out a Common F ou Can TTTTTakakakakake Out a Common F e Out a Common F ou Can YYYYYou Can ou Can actor from Eac e Out a Common Factor fr om Eac actor fr e Out a Common F ou Can If you see a common factor in each term, such as a number or a variable, take it out first. Example Example Example Example Example 11111 Factor 3xΒ² + 15x + 18. Solution Solution Solution Solution Solution 3xΒ² + 15x + 18 = 3(xΒ² + 5x + 6) The expression in parentheses can be factored |
using the method in Topic 6.6.1: = 3(x + 2)(x + 3) But β if the expression in parentheses still has a Ο 1, then the expression will need to be factored using the second method, shown in Example 2. Guided Practice Factor each expression completely. 1. 3x2 + 15x + 12 3. 2t2 β 22t + 60 5. 4x2 + 32x + 64 7. 5m2 + 20m + 15 9. β2k2 β 20k β 32 11. 2 + x β x2 13. 3x2y β 33xy β 126y 15. 100 + 75x β 25x2 17. 32a2 β 8x2a2 19. 2x2m2 + 28x2m β 102x2 2. 4y2 β 12y β 112 4. 3r2 β 75 6. 7p2 + 70p + 63 8. 6x2 + 42x + 60 10. β3m2 β 30m β 72 12. β3x2 β 84x β 225 14. 10x2 + 290x + 1000 16. 100n2y + 100ny β 5600y 18. 21a β 80 β a2 20. 3a2b2x2 + 30a2bx + 63a2 Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 313313313313313 + c b b b b by y y y y TTTTTrial and Er Β² + bx + actor axΒ² + rial and Errrrrrororororor rial and Er rial and Er + + Β² + Β² + actor actor ou Can F YYYYYou Can F ou Can F ou Can Factor rial and Er + Β² + actor ou Can F If you canβt see a common factor, then you need to get axΒ² + bx + c into the form (a1x + c1)(a2x + c2), where a1 and a2 are factors of a, and c1 and c2 are factors of c. a = a1a2 axΒ² + bx + c = (a1x + c1)(a2x + c2) a1c2 + a2c1 = b c = c1c2 Note that |
if a number is positive then its two factors will be either both positive or both negative. If a number is negative, then its two factors will have different signs β one positive and one negative. These facts give important clues about the signs of c1 and c2. Example Example Example Example Example 22222 Factor 3xΒ² + 11x + 6. Solution Solution Solution Solution Solution Write down pairs of factors of a = 3: = 1 and a22222 = a11111 = 1 = 1 3xΒ² + 11x + 6 Write down pairs of factors of c = 6: = 1 and c22222 = c11111 = 1 = 1 = 2 and c22222 = c11111 = 2 = 2 Now find the combination of factors that gives a1c2 + a2c1 = b = 11. Put the x-terms into parentheses first, with the pair of coefficients 3 and 1: (3x )(x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 (and a1c2 β a2c1): (3x 1)(x 6) multiplies to give 18x and x, which add/subtract to give 19x or 17x. (3x 6)(x 1) multiplies to give 3x and 6x, which add/subtract to give 9x or 3x. (3x 2)(x 3) multiplies to give 9x and 2x, which add/subtract to give 1111111111x or 7x. (3x 3)(x 2) multiplies to give 6x and 3x, which add/subtract to give 9x or 3x. So (3x 2)(x 3) is the combination that gives 11x (so b = 11). Now fill in the + / β signs. Both c1 and c2 are positive (since c = c1c2 and b = a1c2 + a2c1 are positive), so the final factors are (3x + 2)(x + 3). Check by expanding the parentheses to make sure they give the original equation: (3x + 2)(x + 3) = 3x2 + 9x + 2x + 6 = 3x2 + 11x + 6 Thatβs what you started with, so (3x + 2)(x + 3) is the correct factorization. |
Check it out: Each pair of coefficients c1 and c2 has TWO possible positions. 314314314314314 Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 Check it out: You can consider separately whether the values of c1 and c2 should be positive or negative. Guided Practice Factor each polynomial. 21. 2x2 + 5x + 2 23. 2y2 + 7y + 3 25. 4x2 + 12x + 9 27. 3x2 + 13x + 12 29. 4a2 + 16a + 7 31. 8a2 + 46a + 11 33. 9x2 + 12x + 4 35. 4b2 + 32b + 55 37. 10a2 + 23a + 12 39. 4x2 + 34x + 16 22. 2a2 + 13a + 11 24. 4x2 + 28x + 49 26. 6x2 + 23x + 7 28. 2x2 + 11x + 5 30. 2p2 + 14p + 12 32. 3g2 + 51g + 216 34. 4a2 + 36a + 81 36. 3x2 + 22x + 24 38. 6t2 + 23t + 20 40. 15b2 + 96b + 36 us Sign us Sign s a Min eful if TTTTTherherherherhereβeβeβeβeβs a Min s a Min eful if eful if Be Car Be Car us Sign s a Minus Sign Be Careful if us Sign s a Min eful if Be Car Be Car Example Example Example Example Example 33333 Factor 6xΒ² + 5x β 6. Solution Solution Solution Solution Solution Write down pairs of factors of a = 6 and a22222 = 1 a11111 = 6 = 1 = 6 = 2 and a22222 = a11111 = 2 = 2 6xΒ² + 5x β 6 Write down pairs of factors of c = β6 (ignoring the minus sign for now): = 1 and c22222 = c11111 = 1 = 1 = 2 and c22222 = c11111 = 2 = 2 Put the x-terms into parentheses first, with the first pair of possible values for a1 and a2, 6 and 1: (6x )(1x ). Now try all the pairs of c1 |
and c2 in the parentheses and find a1c2 + a2c1 and a1c2 β a2c1 like before: (6x 1)(x 6) multiplies to give 36x and x, which add/subtract to give 37x or 35x. (6x 6)(x 1) multiplies to give 6x and 6x, which add/subtract to give 12x or 0x. (6x 2)(x 3) multiplies to give 18x and 2x, which add/subtract to give 20x or 16x. (6x 3)(x 2) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. None of these combinations works, so try again with (2x )(3x ): (2x 1)(3x 6) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. (2x 6)(3x 1) multiplies to give 2x and 18x, which add/subtract to give 20x or 16x. (2x 3)(3x 2) multiplies to give 4x and 9x, which add/subtract to give 13x or 55555x. 5x is what you want, so you can stop there β so (2x 3)(3x 2) is the right combination. Now fill in the + / β signs to get b = +5. One of c1 and c2 must be negative, to give c = β6, so the final factors are either (2x + 3)(3x β 2) or (2x β 3)(3x + 2). The x-term of (2x + 3)(3x β 2) will be 9x β 4x = 5x, whereas the x-term for (2x β 3)(3x + 2) will be 4x β 9x = β5x. So the correct factorization is (2x + 3)(3x β 2). Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 315315315315315 Guided Practice Factor each polynomial. 41. 2x2 + 3x β 2 43. 5k2 + 13k β 6 45. 6b2 β 23b + 7 47. |
3k2 β 2k β 1 49. 18 + 5x β 2x2 51. 9x2 + 12x + 4 53. 3x2 β 7x β 6 55. 6x2 + 2x β 20 57. 6y2 + y β 12 42. 3y2 β y β 2 44. 3x2 β x β 10 46. 2x2 β 5x + 2 48. 3v2 β 16v + 5 50. 28 + x β 2x2 52. 7a2 β 26a β 8 54. 12x2 + 5x β 2 56. 18x2 + x β 4 58. 9m2 β 3m β 20 tic to WWWWWororororork Fk Fk Fk Fk Faster aster aster tic to Use the + and β Signs in the Quadraaaaatic to tic to Use the + and β Signs in the Quadr Use the + and β Signs in the Quadr aster aster tic to Use the + and β Signs in the Quadr Use the + and β Signs in the Quadr Looking carefully at the signs in the quadratic that you are factoring can help to narrow down the choices for a1, a2, c1, and c2. Example Example Example Example Example 44444 Factor 3xΒ² + 11x + 6. Solution Solution Solution Solution Solution Here c = 6, which is positive β so its factors c1 and c2 are either both positive or both negative. But since b = 11 is positive, you can tell that c1 and c2 must be positive (so that a1c2 + a2c1 is positive). Example Example Example Example Example 55555 Factor 3xΒ² β 11x + 6. Solution Solution Solution Solution Solution Here c is also positive, so c1 and c2 are either both positive or both negative. But since b = β11 is negative, you can tell that c1 and c2 must be negative (so that a1c2 + a2c1 is negative). 316316316316316 Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 Example Example Example Example Example 66666 Factor 6xΒ² + 5x β 6. Solution Solution Solution Solution Solution In this expression, c is negative, so one of c1 and c2 must be positive and the other must be negative. So instead of looking at |
both the sums and differences of all the different combinations a1c2 and a2c1, you only need to look at the differences. Guided Practice Factor each expression. 59. 2n2 + n β 3 61. 4a2 + 4a + 1 63. 9y2 + 6y + 1 65. 5x2 β x β 18 67. 6t2 + t β 1 60. 2x2 β 5x β 3 62. 3x2 β 4x + 1 64. 4t2 + t β 3 66. 9x2 β 6x + 1 68. b2 + 10b + 21 Independent Practice Factor each polynomial. 1. 5k2 β 7k + 2 3. 12t2 β 11t + 2 5. 10 + h β 3h2 7. 6 + 5x β 6x2 2. 4k2 β 15k + 9 4. 9 β 7k β 2k2 6. 15x2 β 14x β 8 8. 3x4 + 8x2 β 3 9. If the area of a rectangle is (6x2 + 25x + 14) square units and the length is (3x + 2) units, find the width w in terms of x. 10. The area of a parallelogram is (12x2 + 7x β 10) cm2, where x is positive. If the area is given by the formula Area = base Γ height, find the base length and the height of the parallelogram, given that they are both linear factors of the area. Factor each of these expressions completely. 11. x2ab β 3abx β 18ab 12. (d + 2)x2 β 7x(d + 2) β 18(d + 2) 13. (x + 1)m2 β 2m(x + 1) + (x + 1) 14. β2dk2 β 14dk + 36d 15. Five identical rectangular floor tiles have a total area of (15xΒ² + 10x β 40) mΒ². Find the dimensions of each floor tile, if the length of each side can be written in the form ax + b, where a and b are integers. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Now you can factor lots of different types of polynomials. In the |
next Section youβll learn about another type β quadratic expressions containing two different variables. Section 6.6 Section 6.6 Section 6.6 β Factoring Quadratics Section 6.6 Section 6.6 317317317317317 TTTTTopicopicopicopicopic 6.7.16.7.1 6.7.16.7.1 6.7.1 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to factor quadratic expressions containing two variables. Key words: quadratic factor trial and error Check it out: Following the rules from Topic 6.6.2, both of the terms must be positive β so the + signs can be put in the parentheses from the start. Section 6.7 actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr actoring Quadr in TTTTTwwwwwo o o o o VVVVV |
ariaariaariaariaariabbbbbleslesleslesles in in in in in TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbleslesleslesles in in in in So far, most of the quadratics youβve factored have had only one variable β but the same rules apply if there are two variables. actorededededed actor actor Also Be F Also Be F les Can tics with TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbles Can les Can tics with Quadraaaaatics with tics with Quadr Quadr Also Be Factor les Can Also Be F actor Also Be F les Can tics with Quadr Quadr In Section 6.6 you saw that a quadratic expression such as xΒ² + 2x + 1 can be written as two factors β in this case (x + 1)(x + 1). The same is true of an expression such as xΒ² + 2xy + yΒ² β it can be written as (x + y)(x + y). The method for factoring an expression like this is the same as before: Example Example Example Example Example 11111 Factor the following expression: xΒ² + 4xy + 3yΒ² Solution Solution Solution Solution Solution xΒ² + 4xy + 3yΒ² = (x + )(x + ) To fill the gaps you need two numbers or expressions that will multiply together to make 3yΒ² and add together to make 4y. Try out some sets of numbers or expressions that multiply to make 3yΒ²: 3yΒ² and 1 add together to make 3yΒ² + 1 3y and y add together to make 4y So xΒ² + 4xy + 3yΒ² = (x + 3y)(x + y). Example Example Example Example Example 22222 Factor the following expression: 3pΒ² + 5pq + 2qΒ² Solution Solution Solution Solution Solution 3pΒ² + 5pq + 2qΒ² = (3p + )(p + ) Filling these gaps is a little more complicated. You need two numbers or expressions that will multiply together to make 2qΒ² and, when multiplied by the 3p and p respectively, add together to make 5pq. Try out some sets of numbers or expressions: (3p + q)(p + 2q) β this would give pq-terms of 6pq and pq, which |
add to make 7pq (3p + 2q)(p + q) β this would give pq-terms of 3pq and 2pq, which add to make 5pq 5pq is what you need, so 3pΒ² + 5pq + 2qΒ² = (3p + 2q)(p + q). 318318318318318 Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 Guided Practice Factor each polynomial below. 1. x2 + 3xy + 2y2 3. a2 + 9ab + 18b2 5. d2 + 21md + 20m2 7. x2 β xy β 2y2 9. a2 β ab β 12b2 2. x2 + 14xy + 40y2 4. p2 + 7pq + 12q2 6. k2 + 8pk + 12p2 8. x2 β 2xz β 8z2 10. x2 β 5xy + 6y2 The areas of the rectangles below are the products of two binomials with integer coefficients. Find the possible length and width of each rectangle. 11. Area = (x2 + 3xa + 2a2) ft2 12. Area = (y2 β 9yb + 14b2) in2 13. Area = (3x2 β 4xb + b2) ft2 14. Area = (5a2 β ab β 18b2) m2 15. Area = (4c2 + 4cd + d2) ft2 16. Area = (12x2 β 5xy β 2y2) in2 e More Pe Pe Pe Pe Possibilities ossibilities ossibilities e Mor or is TTTTTricricricricrickkkkky ify ify ify ify if TTTTTherherherherhere are are are are are Mor e Mor or is rial and Errrrrror is or is rial and Er TTTTTrial and Er rial and Er ossibilities ossibilities e Mor or is rial and Er Example Example Example Example Example 33333 Factor the following expression: 2mΒ² β 11mp + 5pΒ² Solution Solution Solution Solution Solution 2mΒ² can be factored into 2m and m, so: 2mΒ² β 11mp + 5pΒ² = (2 |
m β )(m β ) To fill the gaps you need two terms in p that will multiply together to make 5pΒ², and when multiplied by 2m and m respectively, will add together to make β11mp. Try out some sets of parentheses that multiply to make 5pΒ²: (2m β 5p)(m β p) β this would give mp-terms of β2mp and β5mp, which add to make β7mp (2m β p)(m β 5p) β this would give mp-terms of β10mp and βmp, which add to make β11mp The second one gives the β11mp needed, so 2mΒ² β 11mp + 5pΒ² = (2m β p)(m β 5p). Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 319319319319319 Check it out: Both p-terms must be negative, so that the mp-terms add up to a negative term and the p-terms multiply to give a positive pΒ²-term. Example Example Example Example Example 44444 Factor the following expression: 9xΒ² + 6xz β 8zΒ² Solution Solution Solution Solution Solution The zΒ²-term is negative, so one of the z-terms will be negative and the other positive. So that means there are a lot more combinations to try out. 9xΒ² can be made by either 3x Γ 3x or 9x Γ x, and β8zΒ² can be made by any of β2z Γ 4z, 2z Γ β4z, 8z Γ βz, and β8z Γ z. Try out some sets of parentheses: (9x β 4z)(x + 2z) β this would give 18xz and β4xz, which add to make +14xz (9x + 2z)(x β 4z) β this would give β36xz and 2xz, which add to make β34xz (9x β 2z)(x + 4z) β this would give 36xz and β2xz, which add to make +34xz (9x + 4z)(x β 2z) β this would give β18xz and 4xz, which add to make β14xz (3x β 4z)(3x + 2z) β this would give 6 |
xz and β12xz, which add to make β6xz (3x β 2z)(3x + 4z) β this would give 12xz and β6xz, which add to make +6xz You can stop here because +6xz is the expression you are trying to get. So 9xΒ² + 6xz β 8zΒ² = (3x β 2z)(3x + 4z). Guided Practice Factor each of the polynomials below. 17. 2x2 β 5xy β 3y2 19. 3x2 + 17xy + 10y2 21. 3g2 + 7gh + 4h2 23. 4f 2 β 16gf + 15g2 25. 8m2 β 2mh β 15h2 18. 3m2 β 7mp + 2p2 20. 4x2 + 9xy + 2y2 22. 2a2 + 9ab + 9b2 24. 49w2 + 7wz β 6z2 26. 6x2 + 17xy + 12y2 The areas of the rectangles below are the products of two binomials with integer coefficients. Find the possible dimensions of the rectangles. 27. Area = (2m2 + 3mn β 2n2) ft2 28. Area = (a2 + 8ab + 15b2) in.2 29. Area = (3x2 + 10xy + 8y2) m2 30. Area = (15x2 β 29xy + 14y2) ft2 31. Area = (6a2 + 11ab β 10b2) ft2 32. Area = (6c2 + 10cd + 4d2) m2 320320320320320 Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 Independent Practice Factor each of these polynomials. 1. p2 β 7pq + 10q2 3. c2 + 3cd β 40d2 5. y2 β 8xy + 15x2 7. 2r2 + 11rk + 14k2 2. g2 + 4hg β 21h2 4. m2 + 8mn β 20n2 6. 5p2 + 26pq + 5q2 8. 3x2 β 8xy β 3y2 Simplify and factor the following |
polynomials. 9. (x2 + 13x + 8) β (3x2 + 10x β 3) β (x2 + 2x + 6) + (4x2 β 5x β 2) 10. (5x2 + 2x + 4) β (6x2 β 3x + 7) + (4x2 β x β 4) 11. (4x2 β 6xy β 10y2) β (2x2 β 8xy + 2y2) 12. (6x2 + 3xy + 8y2) β (3x2 β 12xy β 10y2) 13. (2t2 β 8tz β 5z2) β (4t2 + 2tz β 15z2) + (5t2 + tz β 40z2) 14. (6x2 β 4xy β 25y2) β (2x2 + 4xy + 25y2) β (2x2 + 4xy β 18y2) The areas of the circles below are products of p and a binomial squared. Find the radius of each circle. 15. Area of circle is (9px2 + 24pxy + 16py2) in.2 16. Area of the circle is (4px2 β 20pax + 25pa2) m2 17. Area of the circle is (9py2 + 48pzy + 64pz2) ft2 18. Area of the circle is (4pm2 + 20pmn + 25pn2) ft2 The areas of the parallelograms below are the products of two binomials with integer coefficients. Find the dimensions of each parallelogram if a = 10 and b = 5. 19. Parallelogram with area (8a2 β 10ab + 3b2) ft2 20. Parallelogram with area (4a2 β 9b2) ft2 21. Parallelogram with area (12a2 + 11ab β 5b2) ft2 22. Parallelogram with area (40a2 β 51ab β 7b2) ft2 Factor the following polynomials. 23. 6a2z2k + 20z2abk + 16b2z2k 24. 36a2b2c β 15ab3c β 6b4c 25. 25x3y2 β 5x2y3 β 90 |
xy4 26. 16a2z2c + 16abz2c + 4b2z2c Factor and simplify completely. 27. 12x2(x + 2) + 25xy(x + 2) + 12y2(x + 2) 28. 18a2b2(a β 1) β 33ab3(a β 1) β 30b4(a β 1) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This is a long process, so itβs easy to make mistakes. You should always check your answer by multiplying out the parentheses again. If you donβt get the expression you started with, you must have gone wrong somewhere. That means youβll need to go back a stage in your work and try a different combination of factors. Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 321321321321321 TTTTTopicopicopicopicopic 6.7.26.7.2 6.7.26.7.2 6.7.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol hniques inc hniques inc tectectectectechniques inc lude finding lude finding hniques include finding lude finding hniques inc lude finding or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms |
in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youβll learn how to factor third-degree expressions. Key words: polynomial degree factor Donβt forget: See Topic 6.6.2 for more on common factors. d-Degggggrrrrreeeeeeeeee d-De d-De actoring TTTTThirhirhirhirhird-De actoring actoring FFFFFactoring d-De actoring actoring TTTTThirhirhirhirhird-De d-Degggggrrrrreeeeeeeeee d-De d-De actoring actoring FFFFFactoring d-De actoring ynomials ynomials PPPPPolololololynomials ynomials ynomials PPPPPolololololynomials ynomials ynomials ynomials ynomials Quadratics are second-degree polynomials because they have an x2-term. Now youβll factor polynomials with an x3-term too. ynomials in Stagggggeseseseses ynomials in Sta d-Degggggrrrrree Pee Pee Pee Pee Polololololynomials in Sta ynomials in Sta d-De actor TTTTThirhirhirhirhird-De d-De actor FFFFFactor actor ynomials in Sta d-De actor To factor a third-degree polynomial, the first thing you should do is look for a common factor. Separate any obvious common factors, then try to factor the remaining expression (the part inside the parentheses). Example Example Example Example Example 11111 Factor xΒ³ + 7xΒ² + 12x completely. Solution Solution Solution Solution Solution All of the terms in xΒ³ + 7xΒ² + 12x contain x, so x is a factor: xΒ³ + 7xΒ² + 12x = x β ββββ β 3 x x + 27 x x + x 12 x = x(x2 + 7x + 12) β ββ |
ββ β Now look at the factor in the parentheses β this is a quadratic expression that it may be possible to factor. In this case itβs possible to factor it, using the method from Section 6.6: xΒ³ + 7xΒ² + 12x = x(x2 + 7x + 12) = x(x + 3)(x + 4) Guided Practice Factor completely these polynomials. 1. 4y3 + 26y2 + 40y 3. 6x3 β 7x2 β 20x 5. 6a3b2 + 33a2b2 + 15ab2 7. 10a3b2c2 + 45a2b2c2 + 20ab2c2 9. 12b4y + 38b3y + 30b2y 11. 36h3k4 + 12h2k4 β 63hk4 13. 189b6c4 β 60b5c4 β 96b4c4 15. 12x4y2 + 78x3y2 + 108x2y2 17. 18b4c β 87b3c + 105b2c 2. 24x3 β 33x2 + 9x 4. 12x3 β 18x2 β 12x 6. 6k3j3 β 10k2j3 β 4kj3 8. 16x4z2 + 12x3z2 β 4x2z2 10. 24c5t2 + 132c4t2 + 144c3t2 12. 40x6y5 + 11x5y5 β 2x4y5 14. 24x2b2c2 + 44xb2c2 β 140b2c2 16. 12a4b2 + 46a3b2 + 40a2b2 18. 12a4b2c β 46a3b2c + 40a2b2c 322322322322322 Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 Example Example Example Example Example 22222 Factor β2xΒ³ β 2xΒ² + 4x completely. Solution Solution Solution Solution Solution Again, x is a factor. Each term has an even coefficient β so you can take out a factor of 2. And given that two of the terms are negative, you can take out a factor of |
β2 instead of 2. This is helpful because it means that the coefficient of xΒ² becomes 1 β which makes the quadratic expression much easier to factor, as you saw in Section 6.6. β2xΒ³ β 2xΒ² + 4x = β2x β ββββ β 2x(x2 + x β 2) Now factor the quadratic: β2xΒ³ β 2xΒ² + 4x = β2x(x2 + x β 2) = β2x(x β 1)(x + 2) β ββββ β Note that in Example 2, if 2x had been factored out instead of β2x, the result would have been: β2xΒ³ β 2xΒ² + 4x = 2x 3 β ββββ β β β ββββ β = 2x(βx2 β x + 2) = 2x(1 β x)(x + 2) or 2x(x β 1)(βx β 2)...which is also correct, but is a little trickier to factor. Guided Practice Factor completely the polynomials below. 19. β4x2y2 β 6xy2 + 4y2 21. β12a3b3 β 30a3b2 + 18a3b 23. β4a3b β 26a2b β 36ab 25. β16a3b2 + 16ab2 27. β147x3y + 84x2y β 12xy 29. 18b4c β 66b3c + 60b2c 31. β42c4d β 28c3d + 14c2d 33. β162a2b3 + 2a2b 35. β60b2d6 + 9b2d5 + 6b2d4 20. β8xy4 + 4xy3 + 60xy2 22. β8w3k β 42w2k β 10wk 24. β90b4c2 β 174b3c2 β 48b2c2 26. β45x4y2 β 50x3y2 β 5x2y2 28. β162c5d + |
288c4d β 128c3d 30. β50y4z β 130y3z + 60y2z 32. β54y3z2 + 21y2z2 + 3yz2 34. 16a4b2 + 176a3b2 + 484a2b2 36. β12b4f2 + 68b3f 2 β 96b2f 2 Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 323323323323323 Independent Practice Factor these polynomials completely. 1. β8a3 + 78a2 + 20a 3. 14y2d2 + 7yd2 β 42d2 5. 250x2y4 + 100x2y3 + 10x2y2 2. 18x2y2 + 51xy2 + 15y2 4. 30x3y3 β 35x2y3 β 100xy3 6. 120a3b3c3 + 28a2b3c3 β 8ab3c3 7. Which of the following is equivalent to 18a2x2 + 3a2x β 3a2? (i) 3a2(2x + 1)(2x β 1) (ii) 3a2(3x β 1)(2x + 1) 8. Which of the following is equivalent to 18a2x2 β 32a2? (i) 2a2(3x + 4)(3x β 4) (ii) 2a2(3x β 4)(3x β 4) 9. Which of the following is a factor of 35x2 + 64ax + 21a2? (i) 5x + 3a (iii) 7x + 5a (ii) 5x + 7a (iv) 7x + 7a 10. Which of the following is a factor of 8y2a2 β 50ya3 β 42a4? (i) 2a3 (iii) 4y β 7a (ii) 8a2 (iv) 4y + 3a Find the value of? in the problems below. (The symbol ββ«β means that the equation is true for all values of the variables.) 11. 81j4 β 36j? + 4j2 β« j2(9j β 2)2 12. 18w3 β 48w? |
+ 32w β« 2w(3w β 4)2 13. 12a3b + 70a?b + 72ab β« 2ab(2a + 9)(3a + 4) 14. β12a4b? β 58a3b3 β 70a2b3 β« β2a2b3(2a + 5)(3a + 7) 15. β56a4b4 + 12a3b4 + 8a2b4 β« β?a2b4(2a β 1)(7a + 2) 16. A cylinder has a base with dimensions that are binomial factors. If the volume of the cylinder is (75px3 + 30px2 + 12px) in.3 and the height is 3x in., find the radius of the base, in terms of x. (V = pr2h) 17. A rectangular prism has a base with dimensions that are binomial factors. If the volume of the prism is (30x3 β 28x2 β 16x) in3 and the height is 2x in., find the dimensions of the base, in terms of x. 18. The product of three consecutive odd integers is x3 + 6x2 + 8x. Find each of the three integers in terms of x. 19. The volume of a rectangular box is (6x3 + 17x2 + 7x) cubic inches, and its height is x inches. Find the dimensions of the base of the box, w and l, in terms of x, given that w and l can be expressed in the form (ax + b), where a and b are integers. 20. A cylinder has a base with dimensions that are binomial factors. If the volume of the cylinder is (36px3 β 96px2 + 64px) ft3 and the height is 4x ft, find the radius of the base, in terms of x. (V = pr2h) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Take care when youβre factoring cubic expressions β you need to take it step by step. Often youβll be able to factor out a term containing x β then youβll be left with a normal quadratic inside the parentheses. Look back at Section 6.6 if youβre having trouble with |
the quadratic part. 324324324324324 Section 6.7 Section 6.7 Section 6.7 β More on Factoring Polynomials Section 6.7 Section 6.7 TTTTTopicopicopicopicopic 6.8.16.8.1 6.8.16.8.1 6.8.1 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- dededededegggggrrrrree pol ynomials..... TTTTThese hese hese ynomials ynomials ee pol ee pol ee polynomials hese hese ynomials ee pol lude lude hniques inc tectectectectechniques inc hniques inc lude finding a hniques include lude hniques inc common factor for all terms in a polynomial, rrrrrecoecoecoecoecognizing gnizing gnizing gnizing gnizing ence of tw tw tw tw twooooo ence of ence of the difffffferererererence of the dif the dif the dif ence of the dif squareseseseses, and recognizing squar squar squar squar perfect squares of binomials..... What it means for you: Youβll use the difference of two squares to factor quadratics. Key words: difference of two squares quadratic Section 6.8 ence of ence of he Difffffferererererence of he Dif he Dif TTTTThe Dif ence of ence of he Dif TTTTThe Dif he Difffffferererererence of ence of ence of he Dif he Dif ence of he Dif ence of o Squareseseseses o Squar o Squar TTTTTwwwwwo Squar o Squar TTTTTwwwwwo Squar o Squareseseseses o Squar o Squar o Squar |
Being able to recognize the difference of two squares is really useful β it helps you factor quadratic expressions. actor Quadraaaaaticsticsticsticstics actor Quadr actor Quadr es to F es to F o Squar ence of TTTTTwwwwwo Squar o Squar ence of Use Difffffferererererence of ence of Use Dif Use Dif es to Factor Quadr o Squares to F actor Quadr es to F o Squar ence of Use Dif Use Dif A difference of two squares is one term squared minus another term squared: m2 β c2. You can use this equation to factor the difference of two squares: m2 β c2 = (m + c)(m β c) The difference of two squares m2 β c2 = The sum of the two terms (m + c) Γ The difference (m β c) between the two terms Example Example Example Example Example 11111 Factor x2 β 9. Donβt forget: The difference of two squares equation was derived in Topic 6.4.1. Solution Solution Solution Solution Solution Substitute x2 for m2 and 9 for c2 in the difference of two squares equation and you get: x2 β 9 = (x + 3)(x β 3) The square root of 9 is 3. Donβt forget the opposite signs. Guided Practice Factor each expression completely. 1. x2 β 16 3. c2 β 49 5. x2 β y2 7. 64 β c2 9. 11x2 β 176 11. 7x2 β 63 13. 3m2n β 12n 15. 162a β 2a3 2. a2 β 25 4. a2 β 100 6. 81 β x2 8. 144 β y2 10. 3y2 β 300 12. 5a2 β 125 14. 6a3 β 216a 16. 7x3y β 7xy3 Section 6.8 Section 6.8 Section 6.8 β More on Quadratics Section 6.8 Section 6.8 325325325325325 Each h h h h TTTTTererererermmmmm Eac Eac oot of k Out the Square Re Re Re Re Root of oot of k Out the Squar WWWWWororororork Out the Squar k Out the Squar oot of Eac Eac oot of k Out the Squar |
Example Example Example Example Example 22222 Factor 4x2 β 25b2. Solution Solution Solution Solution Solution 4x2 = (2x)2, so the square root of 4x2 is 2x. 25b2 = (5b)2, so the square root of 25b2 is 5b. Put the values into the difference of two squares equation: m2 β c2 = (m + c)(m β c) 4x2 β 25b2 = (2x + 5b)(2x β 5b) Square root of 4x2 Square root of 25b2 Guided Practice Factor each expression completely. 17. a2 β 4b2 19. 9a2 β 64x4 21. 49x2 β 4y2 23. 2c3 β 98c 25. 36x2b2 β 49a2c2 27. 75m3n2 β 108a2b2m 18. 4y2 β 81x2 20. 4a2 β 16x2 22. 25y4 β 81c8 24. 3m2n3 β 12a2n 26. 16xbc2 β xba2 28. 18a2b β 242c2b Independent Practice Factor each polynomial completely. 1. 5m3n β 80mn 3. 343m3 β 252mn2 5. 3a3b β 3ab3 7. 4(3x + 2)a2b2 β 9(3x + 2) 9. 3(9a2 β 64b2)a + 8b(9a2 β 64b2) 10. (2x + 5b)4x2 β (2x + 5b)25b2 2. 54a3b β 24ab 4. 50b3 β 18bc2 6. (2x + 1)x2 β (2x + 1) 8. 4a2b2 β 100b2 The areas of the rectangles below are the products of two binomials. Find the two binomials. 11. Area = (81x2 β 100y2) ft2 13. Area = (100a2 β b2) ft2 12. Area = (16a2 β 9y2) ft2 14. Area = (4a2b2 β 25x2) cm2 1 2 (x2 β 8b2) in.2. Find expressions for its base 15. A |