Polynomials: Factoring - XYZ Custom Plus

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748Chapter 12 Polynomials: FactoringGetting Ready for the Next SectionMultiply each of the following.67. (x − 7)(x + 2) 68. (x − 7)(x − 2) 69. (x − 3)(x + 2) 70. (x + 3)(x − 2)71. (x + 3)(x 2 − 3x + 9) 72. (x − 2)(x 2 + 2x + 4) 73. (2x + 1)(x 2 + 4x − 3) 74. (3x + 2)(x 2 − 2x − 4)75. 3x 4 (6x 3 − 4x 2 + 2x) 76. 2x 4 (5x 3 + 4x 2 − 3x) 77. x + 1 _3 x + 2 _3 78. x + 1 _4 x + 3 _4 79. (6x + 4y)(2x − 3y) 80. (8a − 3b)(4a − 5b) 81. (9a + 1)(9a − 1) 82. (7b + 1)(7b + 1)83. (x − 9)(x − 9) 84. (x − 8)(x − 8) 85. (x + 2)(x 2 − 2x + 4) 86. (x − 3)(x 2 + 3x + 9)Maintaining Your SkillsDivide.87. __y 3 − 16y 2 + 64y 88. __5x 3 + 35x 2 + 60x 89. ___−12x 4 + 48x 3 + 144x 2y5x−12x 290. __16x 5 + 20x 4 + 60x 3 91. ___−18y 5 + 63y 4 − 108y 3 92. __36y 6 − 66y 5 + 54y 44x 3 −9y 3 6y 4Subtract.93. (5x 2 + 5x − 4) − (3x 2 − 2x + 7) 94. (7x 4 − 4x 2 − 5) − (2x 4 − 4x 2 + 5) 95. Subtract 4x − 5 from 7x + 3.96. Subtract 3x + 2 from −6x + 1. 97. Subtract 2x 2 − 4x from 5x 2 − 5. 98. Subtract 6x 2 + 3 from 2x 2 − 4x.
12.2Factoring TrinomialsIn this section we will factor trinomials in which the coefficient of the squaredterm is 1. The more familiar we are with multiplication of binomials, the easierfactoring trinomials will be.Recall multiplication of binomials from Chapter 5:(x+ 3)(x+ 4) = x 2 + 7x+ 12(x− 5)(x+ 2) = x 2 − 3x− 10The first term in the answer is the product of the first terms in each binomial.The last term in the answer is the product of the last terms in each binomial. Themiddle term in the answer comes from adding the product of the outside terms tothe product of the inside terms.Let’s have a and b represent real numbers and look at the product of (x+ a)and (x+ b):(x+ a)(x+ b) = x 2 + ax + bx + ab= x 2 + (a + b)x + abThe coefficient of the middle term is the sum of a and b. The last term is the productof a and b. Writing this as a factoring problem, we have:x 2 + (a + b)x + ab = (x+ a)(x+ b)SumProductA Factoring with a Leading Coefficient of 1To factor a trinomial in which the coefficient of x 2 is 1, we need only find the numbersa and b whose sum is the coefficient of the middle term and whose productis the constant term (last term).ExAmplE 1SOlutiOnFactor x 2 + 8x+ 12.The coefficient of x 2 is 1. We need two numbers whose sum is 8 andwhose product is 12. The numbers are 6 and 2:x 2 + 8x+ 12 = (x+ 6)(x+ 2)We can easily check our work by multiplying (x+ 6) and (x+ 2)Check: (x+ 6)(x+ 2) = x 2 + 6x+ 2x + 12= x 2 + 8x+ 12ObjectivesA Factor a trinomial whose leadingcoefficient is the number 1.B Factor a polynomial by firstfactoring out the greatest commonfactor and then factoring thepolynomial that remains.NoteAs you will see as weprogress throughthe book, factoringis a tool that is used in solving anumber of problems. Before seeinghow it is used, however, we firstmust learn how to do it. So, in thissection and the two sections thatfollow, we will be developing ourfactoring skills.prACtiCE prOBlEmS1. Factor x 2 + 5x+ 6.Examples now playing atMathTV.com/booksExAmplE 2Factor x 2 − 2x − 15.SOlutiOn The coefficient of x 2 is again 1. We need to find a pair of numberswhose sum is −2 and whose product is −15. Here are all the possibilities forproducts that are −15.productsSums−1(15) = −15 −1 + 15 = 141(−15) = −15 1 + (−15) = −14−5(3) = −15 −5 + 3 = −25(−3) = −15 5 + (−3) = 22. Factor x 2 + 2x − 15.NoteAgain, we can checkour results by multiplyingour factors tosee if their product is the originalpolynomial.The third line gives us what we want. The factors of x 2 − 2x − 15 are (x− 5)and (x+ 3):x 2 − 2x − 15 = (x− 5)(x+ 3)12.2 Factoring TrinomialsAnswers1. (x+ 2)(x+ 3) 2. (x+ 5)(x− 3)749
Page 1 and 2: Polynomials: Factoring12Introductio
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Chapter 12 TestFactor out the great
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