Section P.6 P.6 Factoring Trinomials 61 Factoring Trinomials What you should learn: • Factor trinomials of the form x2 bx c • Factor trinomials of the form ax2 bx c • Factor trinomials by grouping • Factor perfect square trinomials • Select the best factoring technique using the guidelines for factoring polynomials Why you should learn it: The techniques for factoring trinomials will help you in solving quadratic equations. Factoring Trinomials of the Form x2 bx c Try covering the factored forms in the left-hand column below. Can you determine the factored forms from the trinomial forms? Factored Form F O I L Trinomial Form x 1x 4 4x x 4 x2 3x 4 x 3x 2 x2 2x 3x 6 x2 5x 6 3x 5x 1 3x2 3x 5x 5 3x2 8x 5 x2 Your goal here is to factor trinomials of the form x2 bx c. To begin, consider the factorization x2 bx c x mx n. By multiplying the right-hand side, you obtain the following result. x mx n x2 nx mx mn x2 m nx mn Sum of terms x2 Product of terms b x c So, to factor a trinomial x2 bx c into a product of two binomials, you must find two factors of c with a sum of b. Example 1 Factoring Trinomials Study Tip Use a list to help you find the two numbers with the required product and sum. For Example 1(a): Factors of 8 Sum 1, 8 7 1, 8 7 2, 4 2 2, 4 2 Because 2 is the required sum, the correct factorization is x2 2x 8 x 4x 2. Factor the trinomials (a) x2 2x 8 and (b) x2 5x 6. Solution (a) You need to find two factors whose product is 8 and whose sum is 2. The product of 4 and 2 is 8. x2 2x 8 x 4x 2 The sum of 4 and 2 is 2. (b) You need to find two factors whose product is 6 and whose sum is 5. The product of 3 and 2 is 6 x2 5x 6 x 3x 2 The sum of 3 and 2 is 5. Now try Exercise 7. Note that when the constant term of the trinomial is positive, its factors must have like signs; otherwise, its factors have unlike signs. 62 Chapter P Prerequisites Factors of 24 Sum of Factors 124 124 212 212 38 38 46 46 1 24 23 1 24 23 2 12 10 2 12 10 3 8 5 3 8 5 4 6 2 4 6 2 Study Tip With any factoring problem, remember that you can check your result by multiplying. For instance, in Example 2, you can check the result by multiplying x 9 by x 2 to see that you obtain x2 7x 18. Remember that not all trinomials are factorable using integers. For instance, x2 2x 4 is not factorable using integers because there is no pair of factors of 4 whose sum is 2. When factoring a trinomial of the form x2 bx c, if you have trouble finding two factors of c with a sum of b, it may be helpful to list all of the distinct pairs of factors and then choose the appropriate pair from the list. For instance, consider the trinomial x2 5x 24. For this trinomial, c 24 and b 5. So, you need to find two factors of 24 with a sum of 5, as shown at the left. With experience, you will be able to narrow this list down mentally to only two or three possibilities whose sums can then be tested to determine the correct factorization, which is x2 5x 24 x 3x 8. Example 2 Factoring a Trinomial Factor the trinomial x2 7x 18. Solution To factor this trinomial, you need to find two factors whose product is 18 and whose sum is 7. The product of 2 and 9 is 18. x2 7x 18 x 2x 9 The sum of 2 and 9 is 7. Now try Exercise 11. Applications of algebra sometimes involve trinomials that have a common monomial factor. To factor such trinomials completely, first factor out the common monomial factor. Then try to factor the resulting trinomial by the methods given in this section. Example 3 Factoring Completely Factor the trinomials (a) 4x3 8x2 60x and (b) 5x2y 20xy2 15y3 completely. Solution (a) This trinomial has a common monomial factor of 4x. So, you should start the factoring process by factoring 4x out of each term. 4x3 8x2 60x 4xx2 2x 15 4xx x 4xx 3x 5 Factor out common monomial factor. Think: You need two factors of 15 with a sum of 2. 35 15, 3 5 2 (b) This trinomial has a common monomial factor of 5y. So, you should start the factoring process by factoring 5y out of each term. 5yx x Factor out common monomial factor. Think: You need two factors of 3 with a sum of 4. 5yx yx 3y 13 3, 1 3 4 5x2y 20xy2 15y3 5yx2 4xy 3y2 Now try Exercise 19. Section P.6 Factoring Trinomials 63 Factoring Trinomials of the Form ax2 bx c To factor a trinomial whose leading coefficient is not 1, use the following pattern. Factors of a ax2 bx c x x Factors of c The goal is to find a combination of factors of a and c such that the outer and inner products add up to the middle term bx. For instance, in the trinomial 6x2 17x 5, a 6, c 5, and b 17. After some experimentation, you can determine that the factorization is 6x2 17x 5 2x 53x 1. Example 4 Factoring a Trinomial of the Form ax2 bx c Factor the trinomial 6x2 5x 4. Solution First, observe that 6x2 5x 4 has no common monomial factor. For this trinomial, you have ax2 bx c 6x2 5x 4, which implies that a 6, c 4, and b 5. The possible factors of the leading coefficient 6 are 16 and 23, and the possible factors of 4 are 14, 14, and 22. By trying the many different combinations of these factors, you obtain the following list. Study Tip If the original trinomial has no common monomial factor, its binomial factors cannot have common monomial factors. So, in Example 4, you do not have to test factors, such as 6x 4, that have a common factor of 2. Which of the other factors in Example 4 did not need to be tested? x 16x 4 6x2 2x 4 x 16x 4 6x2 2x 4 x 46x 1 6x2 23x 4 x 46x 1 6x2 23x 4 x 26x 2 6x2 10x 4 x 26x 2 6x2 10x 4 2x 13x 4 6x2 5x 4 2x 13x 4 6x2 5x 4 2x 43x 1 6x2 10x 4 2x 43x 1 6x2 10x 4 2x 23x 2 6x2 2x 4 2x 23x 2 6x2 2x 4 Correct factorization So, you can conclude that the correct factorization is 6x2 5x 4 2x 13x 4. Check this result by multiplying 2x 1 by 3x 4. Now try Exercise 33. 64 Chapter P Prerequisites To help shorten the list of possible factorizations of a trinomial of the form ax2 bx c, use the guidelines presented below. Guidelines for Limiting Possible Trinomial Factorizations 1. If the trinomial has a common monomial factor, you should factor out the monomial factor before trying to find binomial factors. For instance, the trinomial 12x2 10x 8 has a common factor of 2. By removing this common factor, you obtain 12x2 10x 8 26x2 5x 4. 2. Do not switch the signs of the factors of c unless the middle term is correct except in sign. In Example 4, after determining that x 46x 1 is not the correct factorization, for instance, it is unnecessary to test x 46x 1. 3. Do not use binomial factors that have a common monomial factor. Such a factor cannot be correct, because the trinomial has no common monomial factor. (Any common monomial factor should already have been factored out, in accordance with Guideline 1.) For instance, in Example 4, it is unnecessary to test x 16x 4 6x2 2x 4 because the factor 6x 4 has a common factor of 2. Using these guidelines, you could shorten the list given in Example 4 to the following three possible factorizations. x 46x 1 6x2 23x 4 2x 13x 4 6x2 5x 4 2x 13x 4 6x2 5x 4 Correct factorization Do you see why you can cut the list from 12 possible factorizations to only three? Example 5 Factoring a Trinomial of the Form ax2 bx c Factor the trinomial 2x2 x 21. Solution First observe that 2x2 x 21 has no common monomial factor. For this trinomial, a 2, which factors as 12, and c 21, which factors as 121, 121, 37, or 73. 2x 1x 21 2x2 41x 21 2x 21x 1 2x2 19x 21 2x 3x 7 2x2 11x 21 2x 7x 3 2x2 x 21 2x 7x 3 2x2 x 21 Middle term has incorrect sign. Correct factorization So, the correct factorization of x 21 is 2x 7x 3. Check this result by multiplying 2x 7 by x 3. 2x2 Now try Exercise 35. Remember that if the middle term is correct except in sign, you need only change the signs of the factors of c, as in Example 5. Section P.6 Factoring Trinomials 65 Example 6 Factoring Trinomials Factor the trinomials (a) 3x2 11x 10 and (b) 5x2 9xy 4y2. Solution (a) First, observe that 3x2 11x 10 has no common monomial factor. For this trinomial, a 3, which factors as 13, and c 10, which factors as 110 or 25. You can test the possible factors as follows. x 103x 1 3x2 31x 10 x 13x 10 3x2 13x 10 x 53x 2 3x2 17x 10 x 23x 5 3x2 11x 10 Correct factorization So, the correct factorization is 3x2 11x 10 x 23x 5. (b) First observe that 5x2 9xy 4y2 has no common monomial factor. For this trinomial, a 5, which factors as 15, and c 4, which factors as 14 or 22. You can test the possible factors as follows. x 4y5x y 5x2 21xy 4y2 x y5x 4y 5x2 9xy 4y2 Correct factorization So, the correct factorization is 5x2 9xy 4y2 x y5x 4y. Now try Exercise 37. Remember that if a trinomial has a common monomial factor, the common monomial factor should be removed first. This is illustrated in the next two examples. Example 7 Factoring Completely Factor 8x2 y 60xy 28y completely. Solution Begin by factoring out the common monomial factor 4y. 8x2 y 60xy 28y 4y2x2 15x 7 Now, for the new trinomial 2x2 15x 7, a 2 and c 7. The possible factorizations of this trinomial are as follows. 2x 7x 1 2x2 9x 7 2x 1x 7 2x2 15x 7 Correct factorization So, the complete factorization of the original trinomial is 8x2 y 60xy 28y 4y2x2 15x 7 4y2x 1x 7. Now try Exercise 49. 66 Chapter P Prerequisites Example 8 Factoring Completely Factor 4x4 2x3y 6x2y2 completely. Solution Begin by factoring out the common monomial factor 2x2. 4x4 2x3y 6x2y2 2x22x2 xy 3y2 Now, for the new trinomial 2x2 xy 3y2, a 2, and c 3. The possible factorizations of this trinomial are as follows. Study Tip When factoring a trinomial with a negative leading coefficient, first factor 1 out of the trinomial, as demonstrated in Example 9. 2x yx 3y 2x2 5xy 3y2 2x 3yx y 2x2 xy 3y2 2x 3yx y 2x2 xy 3y2 Transpose lines. Middle term has incorrect sign. Correct factorization So, the complete factorization of the original trinomial is 4x 4 2x3y 6x2y2 2x22x2 xy 3y2 2x22x 3yx y. Now try Exercise 51. Example 9 A Trinomial with a Negative Leading Coefficient Factor the trinomial 3x2 16x 35. Solution This trinomial has a negative leading coefficient, so you should begin by factoring 1 out of the trinomial. 3x2 16x 35 13x2 16x 35 Now, for the new trinomial 3x2 16x 35, you have a 3 and c 35. The possible factorizations of this trinomial are as follows. 3x 1x 35 3x2 104x 35 3x 35x 1 3x2 32x 35 3x 5x 7 3x2 16x 35 3x 5x 7 3x2 16x 35 Middle term has incorrect sign. Correct factorization So, the correct factorization is 3x2 16x 35 13x 5x 7. Alternative forms of this factorization include 3x 5x 7 and 3x 5x 7. Now try Exercise 53. Not all trinomials are factorable using only integers. For instance, to factor x2 3x 5 you need factors of 5 that add up to 3. This is not possible, because the only integer factors of 5 are 1 and 5, and their sum is not 3. Such a trinomial is not factorable over the integers. Polynomials that cannot be factored using integer coefficients are called prime with respect to the integers. Some other examples of prime polynomials are 2x2 3x 2, 4x2 9, and 2x2 xy 7y2. Watch for other trinomials that are not factorable in the exercises for this section. Section P.6 Factoring Trinomials 67 Factoring Trinomials by Grouping (Optional) In this section, you have seen that factoring a trinomial can involve quite a bit of trial and error. An alternative technique that some people like to use is factoring by grouping. For instance, suppose you rewrite the trinomial 2x2 x 15 as 2x2 x 15 2x2 6x 5x 15. Then, by grouping the first and second terms and the third and fourth terms, you can factor the polynomial as follows. 2x2 x 15 2x2 6x 5x 15 2x2 6x 5x 15 2xx 3 5x 3 x 32x 5 Rewrite middle term. Group terms. Factor out common monomial factor in each group. Distributive Property The key to this method of factoring is knowing how to rewrite the middle term. In general, to factor a trinomial ax2 bx c by grouping, choose factors of the product ac that add up to b and use these factors to rewrite the middle term. This technique is illustrated in Example 10. Study Tip You should put a polynomial in standard form before trying to factor by grouping. Then group and remove a common monomial factor from the first two terms and the last two terms. Finally, if possible, factor out the common binomial factor. Example 10 Factoring a Trinomial by Grouping Use factoring by grouping to factor the trinomials completely. (a) 2x2 5x 3 (b) 6y2 5y 4 Solution (a) In the trinomial 2x2 5x 3, a 2 and c 3, which implies that the product ac is 6. Now, because 6 factors as 61, and 6 1 5 b, you can rewrite the middle term as 5x 6x x. This produces the following result. 2x2 5x 3 2x2 6x x 3 2x2 6x x 3 2xx 3 x 3 x 32x 1 Rewrite middle term. Group terms. Factor out common monomial factor in first group. Distributive Property So, the trinomial factors as 2x2 5x 3 x 32x 1. (b) In the trinomial 6y2 5y 4, a 6 and c 4, which implies that the product of ac is 24. Now, because 24 factors as 83, and 8 3 5 b, you can rewrite the middle term as 5y 8y 3y. This produces the following result. 6y2 5y 4 6y2 8y 3y 4 6y2 8y 3y 4 2y3y 4 3y 4 3y 42y 1 Rewrite middle term. Group terms. Factor out common monomial factor in first group. Distributive Property So, the trinomial factors as 6y2 5y 4 3y 42y 1. Now try Exercise 59. 68 Chapter P Prerequisites Factoring Perfect Square Trinomials A perfect square trinomial is the square of a binomial. For instance, x2 6x 9 x 3x 3 x 32 is the square of the binomial x 3, and 4x2 20x 25 2x 52x 5 2x 52 is the square of the binomial 2x 5. Perfect square trinomials come in two forms: one in which the middle term is positive, and the other in which the middle term is negative. Perfect Square Trinomials Let u and v represent real numbers, variables, or algebraic expressions. Then the perfect square trinomials below can be factored as indicated. u2 2uv v2 u v2 u2 2uv v2 u v2 Like signs Like signs To recognize a perfect square trinomial, remember that the first and last terms must be perfect squares and positive, and that the middle term must be twice the product of u and v. (The middle term can be positive or negative.) Example 11 Factoring Perfect Square Trinomials Rewrite each trinomial as a perfect square trinomial. Then factor the trinomial. (a) x2 4x 4 (b) 16y2 24y 9 (c) 9x2 30xy 25y2 Solution (a) x2 4x 4 x2 2x2 22 x 22 (b) 16y2 24y 9 4y2 24y3 32 4y 32 (c) 9x2 30xy 25y2 3x2 23x5y 5y2 3x 5y2 Now try Exercise 63. Example 12 Factoring out a Common Monomial Factor First Factor out any common monomial factors from each trinomial. Then factor completely. (a) 3x2 30x 75 (b) 16y 3 80y 2 100y Solution (a) 3x2 30x 75 3x2 10x 25 3x 52 (b) 16y3 80y2 100y 4y4y2 20y 25 4y2y 52 Now try Exercise 83. Factor out common monomial factor. Factor as perfect square trinomial. Factor out common monomial factor. Factor as perfect square trinomial. Section P.6 Factoring Trinomials 69 Summary of Factoring Although the basic factoring techniques have been discussed one at a time, from this point on you must decide which technique to apply to any given problem situation. The guidelines below should assist you in this selection process. Guidelines for Factoring Polynomials 1. Factor out any common factors. 2. Factor according to one of the special polynomial forms: difference of squares, sum or difference of cubes, or perfect square trinomial. 3. Factor trinomials, ax2 bx c, using the methods for a 1 or a 1. 4. Factor by grouping—for polynomials with four terms. 5. Check to see if the factors themselves can be factored further. 6. Check the results by multiplying the factors. Example 13 Factoring Polynomials Factor each polynomial completely. (a) 3x2 108 (b) 4x3 32x2 64x (d) x3 3x2 4x 12 (c) 6x3 27x2 15x (e) x2 6x 9 y2 Solution (a) 3x2 108 3x2 36 Factor out common factor. 3x 6x 6 (b) 4x3 32x2 64x 4x x2 Difference of two squares 8x 16 4xx 4 2 Perfect square trinomial (c) 6x3 27x2 15x 3x2x2 9x 5 3x2x 1x 5 (d) x3 3x2 4x 12 x3 3x2 4x 12 x 3 4x 3 x 3x2 4 x 3x 2x 2 x2 (e) x2 6x 9 y2 x2 6x 9 y2 x 3 2 Factor out common factor. y2 Factor out common factor. Factor trinomial. Group terms. Factor out common factors. Distributive Property Difference of two squares Group terms. Perfect square trinomial x 3 yx 3 y Difference of two squares x 3 yx 3 y Simplify. Now try Exercise 97. Notice the grouping in the first step of Example 13(e). The first three terms form a perfect square trinomial, which enables you to further factor the polynomial as the difference of two squares. 70 Chapter P P.6 Prerequisites Exercises VOCABULARY CHECK: Fill in the blanks. 1. Polynomials that cannot be factored using integer coefficients are called ________. 2. If a polynomial has four terms, a method of factoring called ________ may be used. 3. A ________ is the square of a binomial. In Exercises 1–6, fill in the missing factor. 1. x2 5x 4 x 4 43. 20x2 x 12 44. 10x2 9xy 9y2 45. 2u2 9uv 35v2 46. 5r 2 4rs 9s2 2. a2 2a 8 a 4 47. 15x2 3xy 8y2 48. 9u2 5uv 6v2 4. y2 6y 8 y 4 49. 6x2y 22xy 40y 50. 6a2b2 57ab2 27b2 51. 8x 4 6x3y 2x2y2 52. 24u3 54u2v 27uv2 3. y2 y 20 y 4 5. z2 6z 8 z 4 6. z2 2z 24 z 4 In Exercises 7–22, factor the trinomial. 7. x2 4x 3 8. x2 10x 24 9. y2 7y 30 10. m2 3m 10 11. t2 4t 21 12. x2 4x 12 13. x2 12x 20 14. y2 8y 33 15. t2 17t 60 17. x2 20x 96 16. z2 13z 36 19. u2 5uv 6v2 21. x2 2xy 35y2 In Exercises 53–56, factor the trinomial by first factoring 1 from each term. 7xy 20. 2 22. a 21ab 110b2 12y2 In Exercises 23–28, fill in the missing factor. 26. 5c2 11c 12 c 3 27. 2y2 3y 27 y 3 28. 3y2 y 30 y 3 In Exercises 29–52, factor the trinomial, if possible. Some of the expressions are not factorable using integer coefficients. 31. 8t2 6t 5 33. 6b2 19b 7 30. 5x2 7x 2 32. 2x2 6x 2 37. 2x2 5xy 3y2 38. 5a2 8ab 3b2 39. 11y2 43yz 4z2 40. 6u2 7uv 5v2 41. ab 42. 59. 6x2 x 2 60. 6x2 x 15 61. 15x2 11x 2 62. 12x2 28x 15 24x2 14xy a2 12a 36 66. y2 14y 49 68. 4z2 28z 49 69. 9b2 12b 4 70. 4x2 4x 1 71. 4x2 4xy y2 72. m2 6mn 9n2 73. u2 8uv 16v2 74. 4y2 20yz 25z2 In Exercises 75–112, completely factor the algebraic expression. 77. 10t3 2t2 36t 3y2 64. z2 6z 9 67. 25y2 10y 1 76. 20y2 45 35. 2a2 13a 20 2b2 58. 2x2 9x 9 75. 3x5 12x3 34. 12a2 2a 4 36. 9t2 51t 30 6a2 57. 3x2 10x 8 65. 12a 9 a 3 29. 3x2 4x 1 56. 2 5x 12x2 63. x2 4x 4 24. 5x2 19x 12 x 3 25. 55. 1 11x 60x2 In Exercises 63–74, factor the perfect square trinomial. 23. 5x2 18x 9 x 3 5a2 54. 6x2 5x 6 In Exercises 57–62, factor the trinomial by grouping. 18. y2 22y 96 x2 53. 2x2 x 6 78. 3x2 12x 36 79. 5a2 25a 30 80. 4t2 8t 32 Section P.6 81. 6u2 3u 63 82. 10x2 83. 9y2 22x 24 In Exercises 119–124, find a real number c such that the algebraic expression is a perfect square trinomial. 66y 121 119. x2 8x c 84. 16z2 56z 49 121. y2 6y c 85. 4x3x 2 3x 22 122. z2 20z c 86. 2x 52 x2x 5 123. 16a2 40a c 87. x2x 3 5xx 3 124. 9t2 12t c 88. 6y32y 5 3y22y 5 89. 36 z 32 90. x 42 25 91. 2y 12 9 92. 81 3t 22 71 Factoring Trinomials 120. x2 12x c In Exercises 125 and 126, fill in the missing number. 125. x2 12x 50 x 62 126. x2 10x 22 x 52 93. 54x3 2 In Exercises 127–132, find all integers b such that the trinomial can be factored. 94. 3t3 24 127. x2 bx 18 95. v 3v 5v 3 2 10a 96. 97. 2x3y 2x2y2 84xy3 2a3 8a2 98. 8m3n 20m2n2 48mn3 99. 5x2y 20y3 100. 27a3 3ab2 101. x3 2x2 16x 32 102. x3 7x2 4x 28 103. x3 6x2 9x 54 104. x3 10x2 16x 160 105. x2 10x 25 y2 106. 9y2 12y 4 z2 107. a2 2ab b2 16 128. x2 bx 14 129. x2 bx 21 130. x2 bx 7 131. 5x2 bx 8 132. 3x2 bx 10 In Exercises 133–138, find two integers c such that the trinomial can be factored. (There are many correct answers.) 133. x2 6x c 134. x2 9x c 135. x2 3x c 136. x2 12x c 137. t2 4t c 108. x2 14xy 49y2 4a2 138. s2 s c 109. x8 1 Geometry 111. b4 216b In Exercises 139–142, write, in factored form, an expression for the area of the shaded portion of the figure. 112. x4 16y4 139. 110. 3y3 192 x 116. 36z2 bz 1 117. 4x2 bx 9 118. 16x2 bxy 25y2 4 x 8 xx x x x 113. x2 bx 81 115. 9y2 by 1 x x In Exercises 113–118, find two real numbers b such that the algebraic expression is a perfect square trinomial. 114. x2 bx 16 140. x 18 9 141. 142. 3 x 6 x+3 4 2x 5 5 (x 4 + 3) 72 Chapter P Prerequisites 143. Geometry A rectangle has area and width as indicated in the figure. Factor to find the length. (d) a a b b 2x − 5 A = 2x 2 + x − 15 1 1 ? 1 1 b 144. Geometry A circle has area as indicated in the figure. Factor to find the radius. 1 145. a2 b2 a ba b 146. a2 2ab b2 a b2 147. a2 2a 1 a 12 ? 148. ab a b 1 a 1b 1 Area = π x 2 − 12 π x + 36π Geometric Modeling In Exercises 149 and 150, factor the trinomial and represent the result with a geometric factoring model. Geometric Modeling In Exercises 145–148, match the factoring formula with the correct geometric factoring model. [ The models are labeled (a), (b), (c), and (d).] (a) a a a 1 149. x2 4x 3 Synthesis 151. Error Analysis a 150. x2 5x 4 Identify the error. 9x2 9x 54 3x 63x 9 1 1 1 (b) 1 a a a b In Exercises 153 and 154, explain how the same result could be obtained mentally using the sample below as a model. b 292 30 12 302 2301 12 b b 152. Think About It Is xx 2 2x 2 completely factored? Explain. Writing a b 900 60 1 b a (c) 3x 2x 3 1 a 841 b 153. 522 2704 a−b a 154. 392 1521 155. (a) Completely factor the expression b b 4x3 36x2 80x. (b) An open box is to be made from a rectangular piece of material 10 inches long and 8 inches wide by cutting equal squares of side x from each corner and turning up the sides. Show that the volume of the box can be described by the factored expression in part (a).
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