Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 303: 26



Work Step by Step

Factoring the $GCF=z$ of the given expression, $ 2x^2z+5xz-12z $, results to \begin{array}{l}\require{cancel} z(2x^2+5x-12) .\end{array} The two numbers whose product is $ac= 2(-12)=-24 $ and whose sum is $b= 5 $ are $\{ -3,8 \}$. Using these two numbers to decompose the middle term, then the factored form of the resulting expression, $ z(2x^2+5x-12) $,is \begin{array}{l}\require{cancel} z(2x^2-3x+8x-12) \\\\= z[(2x^2-3x)+(8x-12)] \\\\= z[x(2x-3)+4(2x-3)] \\\\= z[(2x-3)(x+4)] \\\\= z(2x-3)(x+4) .\end{array}
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