College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises - Page 40: 80



Work Step by Step

Let $a=4z-3.$ Then the given expression, $ 6(4z-3)^2+7(4z-3)-3 ,$ is equivalent to \begin{array}{l}\require{cancel} 6a^2+7a-3 .\end{array} The two numbers whose product is $ac= 6(-3)=-18 $ and whose sum is $b= 7 $ are $\{ 9,-2 \}.$ Using these two numbers to decompose the middle term, then the factored form of $ 6a^2+7a-3 $ is \begin{array}{l}\require{cancel} 6a^2+9a-2a-3 \\\\= (6a^2+9a)-(2a+3) \\\\= 3a(2a+3)-(2a+3) \\\\= (2a+3)(3a-1) .\end{array} Since $a=4z-3$, then the expression $ (2a+3)(3a-1) $ is equivalent to \begin{array}{l}\require{cancel} (2(4z-3)+3)(3(4z-3)-1) \\\\= (8z-6+3)(12z-9-1) \\\\= (8z-3)(12z-10) \\\\= (8z-3)2(6z-5) \\\\= 2(8z-3)(6z-5) .\end{array}
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