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: 23

Answer

$2(3x-5)(2x+5)$

Work Step by Step

Factoring the $GCF=2$ of the given expression, $ 12x^2+10x-50 $, results to \begin{array}{l}\require{cancel} 2(6x^2+5x-25) .\end{array} The two numbers whose product is $ac= 6(-25)=-150 $ and whose sum is $b= 5 $ are $\{ -10,15 \}$. Using these two numbers to decompose the middle term, then the factored form of the resulting expression, $ 2(6x^2+5x-25) $,is \begin{array}{l}\require{cancel} 2(6x^2-10x+15x-25) \\\\= 2[(6x^2-10x)+(15x-25)] \\\\= 2[2x(3x-5)+5(3x-5)] \\\\= 2[(3x-5)(2x+5)] \\\\= 2(3x-5)(2x+5) .\end{array}
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