DIVISION OF POLYNOMIALS

5.5 Division of Polynomials (5-33) 89 5.5 DIVISION OF POLYNOMIALS In this section Dividing a Polynomial by a Monomial Dividing a Polynomial by a Binomial Synthetic Division Division and Factoring We began our study of polynomials in Section 5.3 by learning how to add, subtract, and multiply polynomials. In this section we will study division of polynomials. Dividing a Polynomial by a Monomial You learned how to divide monomials in Section 5.1. For eample, 6 3 (3) 6 3. 3 We check by multiplying. Because 3 6 3, this answer is correct. Recall that a b c if and only if c b a. We call a the dividend, b the divisor, and c the quotient. We may also refer to a b and a b as quotients. We can use the distributive property to find that 3( 5 4) 6 3 15 1. So if we divide 6 3 15 1 by the monomial 3, we must get 5 4. We can perform this division by dividing 3 into each term of 6 3 15 1: 6 3 15 3 1 6 3 3 1 5 3 1 3 5 4 In this case the divisor is 3, the dividend is 6 3 15 1, and the quotient is 5 4. E X A M P L E 1 helpful hint Recall that the order of operations gives multiplication and division an equal ranking and says to do them in order from left to right. So without parentheses, 1 5 3 actually means 1 5 3. Dividing polynomials Find the quotient. a) 1 5 ( 3 ) b) (0 6 8 4 4 ) (4 ) a) When dividing 5 by 3, we subtract the eponents: 1 5 ( 3 ) 1 5 3 6 The quotient is 6. Check: 6 3 1 5 b) Divide each term of 0 6 8 4 4 by 4 : 0 6 8 4 4 4 4 0 6 8 4 4 4 4 5 4 1 The quotient is 5 4 1. Check: 4 (5 4 1) 0 6 8 4 4

90 (5-34) Chapter 5 Eponents and Polynomials Dividing a Polynomial by a Binomial We can multiply and 5 to get ( )( 5) 3 10. So if we divide 3 10 by the factor, we should get the other factor 5. This division is not done like division by a monomial; it is done like long division of whole numbers. We get the first term of the quotient by dividing the first term of into the first term of 3 10. Divide by to get. 3 10 Multiply: ( ). 5 Subtract: 3 () 5. Now bring down 10. We get the second term of the quotient (below) by dividing the first term of into the first term of 5 10. Divide 5 by to get 5. 5 5 5 3 10 5 10 5 10 Multiply: 5( ) 5 10. 0 Subtract: 10 (10) 0. So the quotient is 5 and the remainder is 0. If the remainder is not 0, then dividend (divisor)(quotient) (remainder). If we divide each side of this equation by the divisor, we get divide divis nd or quotient re mainder d. ivisor When dividing polynomials, we must write the terms of the divisor and the dividend in descending order of the eponents. If any terms are missing, as in the net eample, we insert terms with a coefficient of 0 as placeholders. When dividing polynomials, we stop the process when the degree of the remainder is smaller than the degree of the divisor. E X A M P L E helpful hint Students usually have the most difficulty with the subtraction part of long division. So pay particular attention to that step and double check your work. Dividing polynomials Find the quotient and remainder for (3 4 5) ( 3). Rearrange 3 4 5 as 3 4 5 and insert the terms 0 3 and 0 : 3 9 7 33 4 0 3 0 5 3 4 9 3 9 3 0 0 3 (9 3 ) 9 3 9 3 7 7 5 7 81 76

5.5 Division of Polynomials (5-35) 91 The quotient is 3 9 7, and the remainder is 76. Note that the degree of the remainder is 1, and the degree of the divisor is. To check, verify that ( 3)(3 9 7) 76 3 4 5. E X A M P L E 3 study tip Have you ever used ecuses to avoid studying? ( Before I can study, I have to do my laundry and go to the bank. ) Since the average attention span for one task is approimately 0 minutes, it is better to take breaks from studying to run errands and to do laundry than to get everything done before you start studying. Rewriting a ratio of two polynomials Write 43 9 3 in the form quotient re mainder. divisor Divide 4 3 9by 3. Insert 0 for the missing term. 3 4 34 3 0 9 4 3 () 4 3 6 6 0 (6 ) 6 6 9 8 9 (9) 8 8 1 3 9 (1) 3 Since the quotient is 3 4 and the remainder is 3, we have 4 3 9 3 3 4. 3 3 To check the answer, we must verify that ( 3)( 3 4) 3 4 3 9. Synthetic Division When dividing a polynomial by a binomial of the form c, we can use synthetic division to speed up the process. For synthetic division we write only the essential parts of ordinary division. For eample, to divide 3 5 4 3by, we write only the coefficients of the dividend 1, 5, 4, and 3 in order of descending eponents. From the divisor we use and start with the following arrangement: 1 5 43 (1 3 5 4 3) ( ) Net we bring the first coefficient, 1, straight down: 1 5 43 Bring down 1 We then multiply the 1 by the from the divisor, place the answer under the 5, and then add that column. Using for allows us to add the column rather than subtract as in ordinary division: Multiply 1 5 1 3 4 3 Add

9 (5-36) Chapter 5 Eponents and Polynomials We then repeat the multiply-and-add step for each of the remaining columns: 1 5 4 3 6 4 Multiply 1 3 7 Remainder Quotient From the bottom row we can read the quotient and remainder. Since the degree of the quotient is one less than the degree of the dividend, the quotient is 1 3. The remainder is 7. The strategy for getting the quotient Q() and remainder R by synthetic division can be stated as follows. Strategy for Using Synthetic Division 1. List the coefficients of the polynomial (the dividend).. Be sure to include zeros for any missing terms in the dividend. 3. For dividing by c, place c to the left. 4. Bring the first coefficient down. 5. Multiply by c and add for each column. 6. Read Q() and R from the bottom row. CAUTION Synthetic division is used only for dividing a polynomial by the binomial c, where c is a constant. If the binomial is 7, then c 7. For the binomial 7 we have 7 (7) and c 7. E X A M P L E 4 Using synthetic division Find the quotient and remainder when 4 5 6 9 is divided by. Since (), we use for the divisor. Because 3 is missing in the dividend, use a zero for the coefficient of 3 : 0 5 6 9 4 8 6 0 Multiply 4 3 0 9 Quotient and remainder 4 0 3 5 6 9 Add Because the degree of the dividend is 4, the degree of the quotient is 3. The quotient is 3 4 3, and the remainder is 9. We can also epress the results of this division in the form quotient re mainder d : ivisor 4 5 6 9 3 4 9 3 Division and Factoring To factor a polynomial means to write it as a product of two or more simpler polynomials. If we divide two polynomials and get 0 remainder, then we can write dividend (divisor)(quotient)

5.5 Division of Polynomials (5-37) 93 and we have factored the dividend. The dividend factors as the divisor times the quotient if and only if the remainder is 0. We can use division to help us discover factors of polynomials. To use this idea, however, we must know a factor or a possible factor to use as the divisor. E X A M P L E 5 Using synthetic division to determine factors Is 1 a factor of 6 3 5 4 3? We can use synthetic division to divide 6 3 5 4 3by1: 1 6 5 4 3 6 1 3 6 1 3 0 Because the remainder is 0, 1 is a factor, and 6 3 5 4 3 ( 1)(6 3). E X A M P L E 6 Using division to determine factors Is a b a factor of a 3 b 3? Divide a 3 b 3 by a b. Insert zeros for the missing a b- and ab -terms. a ab b a ba 3 0 0 b 3 a 3 a b a b 0 a b ab ab b 3 ab b 3 0 Because the remainder is 0, a b is a factor, and a 3 b 3 (a b)(a ab b ). WARM-UPS True or false? Eplain your answer. 1. If a b c, then c is the dividend. False. The quotient times the dividend plus the remainder equals the divisor. False 3. ( )( 3) 1 5 7 is true for any value of. True 4. The quotient of ( 5 7) ( 3) is. True 5. If 5 7 is divided by, the remainder is 1. True 6. To divide 3 4 1by3, we use 3 in synthetic division. False 7. We can use synthetic division to divide 3 4 6by 5. False 8. If 3 5 4 3 is divided by, the quotient has degree 4. True 9. If the remainder is zero, then the divisor is a factor of the dividend. True 10. If the remainder is zero, then the quotient is a factor of the dividend. True

94 (5-38) Chapter 5 Eponents and Polynomials 5. 5 EXERCISES Reading and Writing After reading this section, write out the answers to these questions. Use complete sentences. 1. What are the dividend, divisor, and quotient? If a b c, then the divisor is b, the dividend is a, and the quotient is c.. In what form should polynomials be written for long division? For long division polynomials should be written with the eponents in descending order. 3. What do you do about missing terms when dividing polynomials? If the term n is missing in the dividend, insert the term 0 n for the missing term. 4. When do you stop the long division process for dividing polynomials? Stop the long division process when the remainder has a smaller degree than the divisor. 5. What is synthetic division used for? Synthetic division is used only for dividing by a binomial of the form c. 6. What is the relationship between division of polynomials and factoring polynomials? The remainder is zero if and only if the dividend is a factor of the divisor. Find the quotient. See Eample 1. 7. 36 7 (3 3 ) 1 4 8. 30 3 (5) 6 9. 16 (8 ) 10. a 3 (11a ) a 11. (6b 9) 3 b 3 1. (8 6) 4 3 13. (3 6) (3) 14. (5 3 10 0) (5) 4 15. (10 4 8 3 6 ) ( ) 5 4 3 16. (9 3 6 1) (3) 3 4 17. (7 3 4 ) () 7 18. (6 3 5 ) (4 ) 3 5 4 Find the quotient and remainder as in Eample. Check by using the formula dividend (divisor)(quotient) remainder. 19. ( 8 13) ( 3) 5, 0. ( 5 7) ( 3), 1 1. ( ) ( ) 4, 8. (3) ( 1) 3, 3 3. ( 3 8) ( ) 4, 0 4. (y 3 1) (y 1) y y 1, 0 5. (a 3 4a 5) (a ) a a 8, 11 6. (w 3 w 3) (w ) w 3w 6, 9 7. ( 3 3) ( 1) 3, 6 8. (a 3 a a 4) (a ) a 3a 7, 18 9. ( 4 3 1) ( ) 3 3 6 11, 1 30. (3 4 6 3) ( ) 3 3 6 11 19, 44 31. (5 3 4 ) ( ) 3 1, 4 3. ( 4 3 ) ( 3) 3, 3 7 33. ( 4 ) ( 3) 1 5 4, 7 4 34. ( 5 1) (3 6) 1 3 7, 15 3 35. ( 6) (3 ) 3 1 9, 5 6 9 36. (3 4 1) ( 1) 3 5 4, 9 4 Write each epression in the form quotient re mainder. divisor See Eample 3. 37. 38. 5 1 10 1 1 5 1 39. 40. 9 1 3 1 18 1 3 1 3 3 41. 4. 3 1 8 4 7 4 43. 3 44. 3 3 45. 4 9 46. 5 10 3 1 16 6 3 47. 33 4 7 48. 3 3 1 3 6 1 17 5 10 1 Use synthetic division to find the quotient and remainder when the first polynomial is divided by the second. See Eample 4. 49. 3 5 6 3, 3, 3 50. 3 6 3 5, 3 9 4, 67

5.5 Division of Polynomials (5-39) 95 51. 4 5, 1 6, 11 5. 3 7 4, 3 13, 30 53. 3 4 15 7 9, 3 3 3 9 1 43, 10 54. 4 3 5, 3 4 5 10, 5 55. 5 1, 1 4 3 1, 0 56. 6 1, 1 5 4 3 1, 0 57. 3 5 6, 1, 8 58. 3 3 7, 4 4 13, 45 59..3 0.14 0.6, 0.3.3 0.596, 0.7907 60. 1.6 3.5 4.7, 1.8 1.6 6.38, 16.184 For each pair of polynomials, determine whether the first polynomial is a factor of the second. Use synthetic division when possible. See Eamples 5 and 6. 61. 4, 3 11 8 No 6. 4, 3 48 No 63. 4, 3 13 1 Yes 64. 1, 3 3 5 No 65. 3, 3 3 4 6 Yes 66. 3 5, 6 7 6 No 67. 3w 1, 7w 3 1 Yes 68. w 3, 8w 3 7 Yes 69. a 5, a 3 15 Yes 70. a, a 6 64 Yes 71., 4 3 3 6 4 Yes 7. 3, 4 3 4 6 3 Yes Factor each polynomial given that the binomial following each polynomial is a factor of the polynomial. 73. 6 8, 4 ( 4)( ) 74. 3 40, 8 ( 8)( 5) 75. w 3 7, w 3 (w 3)(w 3w 9) 76. w 3 15, w 5 (w 5)(w 5w 5) 77. 3 4 6 4, ( )( ) 78. 3 5 7, 1 ( 7)( 1) 79. z 6z 9, z 3 (z 3)(z 3) 80. 4a 0a 5, a 5 (a 5)(a 5) 81. 6y 5y 1, y 1 (y 1)(3y 1) 8. 1y y 6, 4y 3 (4y 3)(3y ) Solve each problem. 83. Average cost. The total cost in dollars for manufacturing professional racing bicycles in one week is given by the polynomial function C() 0.03 300. Cost (in thousands of dollars) The average cost per bicycle is given by AC() C (). a) Find a formula for AC(). AC() 0.03 300 b) Is AC() a constant function? No c) Why does the average cost look constant in the accompanying figure? Because AC() is very close to 300 for less than 15, the graph looks horizontal. 4 3 1 0 0 C() AC() 5 10 15 Number of bicycles FIGURE FOR EXERCISE 83 84. Average profit. The weekly profit in dollars for manufacturing bicycles is given by the polynomial P() 100. The average profit per bicycle is given ). Find AP(). Find the average profit per by AP() P( bicycle when 1 bicycles are manufactured. AP() 100, $14 85. Area of a poster. The area of a rectangular poster advertising a Pearl Jam concert is 1 square feet. If the length is 1 feet, then what is the width? 1 feet 86. Volume of a bo. The volume of a shipping crate is h 3 5h 6h. If the height is h and the length is h, then what is the width? h 3 h h FIGURE FOR EXERCISE 86 87. Volume of a pyramid. Ancient Egyptian pyramid builders knew that the volume of the truncated pyramid shown in the figure on the net page is given by V H 3 ( a b 3 ), 3( a b)?

96 (5-40) Chapter 5 Eponents and Polynomials where a is the area of the square base, b is the area of the square top, and H is the distance from the base to the top. Find the volume of a truncated pyramid that has a base of 900 square meters, a top of 400 square meters, and a height H of 10 meters. 6,333.3 cubic meters a b FIGURE FOR EXERCISE 87 H b a 88. Egyptian pyramid formula. Rewrite the formula of the previous eercise so that the denominator contains the number 3 only. V H(a ab b ) 3 GETTING MORE INVOLVED 89. Discussion. On a test a student divided 3 3 5 3 7 by 3 and got a quotient of 3 4 and remainder 9 7. Verify that the divisor times the quotient plus the remainder is equal to the dividend. Why was the student s answer incorrect? 90. Eploration. Use synthetic division to find the quotient when 5 1 is divided by 1 and the quotient when 6 1 is divided by 1. Observe the pattern in the first two quotients and then write the quotient for 9 1 divided by 1 without dividing. 5.6 FACTORING POLYNOMIALS In this section Factoring Out the Greatest Common Factor (GCF) Factoring Out the Opposite of the GCF Factoring the Difference of Two Squares Factoring Perfect Square Trinomials Factoring a Difference or Sum of Two Cubes Factoring a Polynomial Completely Factoring by Substitution In Section 5.5 you learned that a polynomial could be factored by using division: If we know one factor of a polynomial, then we can use it as a divisor to obtain the other factor, the quotient. However, this technique is not very practical because the division process can be somewhat tedious, and it is not easy to obtain a factor to use as the divisor. In this section and the net two sections we will develop better techniques for factoring polynomials. These techniques will be used for solving equations and problems in the last section of this chapter. Factoring Out the Greatest Common Factor (GCF) A natural number larger than 1 that has no factors other than itself and 1 is called a prime number. The numbers, 3, 5, 7, 11, 13, 17, 19, 3 are the first nine prime numbers. There are infinitely many prime numbers. To factor a natural number completely means to write it as a product of prime numbers. In factoring 1 we might write 1 4 3. However, 1 is not factored completely as 4 3 because 4 is not a prime. To factor 1 completely, we write 1 3 (or 3). We use the distributive property to multiply a monomial and a binomial: 6( 1) 1 6 If we start with 1 6, we can use the distributive property to get 1 6 6( 1). We have factored out 6, which is a common factor of 1 and 6. We could have factored out just 3 to get 1 6 3(4 ), but this would not be factoring out the greatest common factor. The greatest common factor (GCF) is a monomial that includes every number or variable that is a factor of all of the terms of the polynomial.