Intermediate Algebra (6th Edition)

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

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 10



Work Step by Step

Factoring the $GCF=2$, then the given expression, $ 12x^2-22x-20 $, is equivalent to \begin{array}{l} 2(6x^2-11x-10) .\end{array} The two numbers whose product is $ac= 6(-10)=-60 $ and whose sum is $b= -11 $ are $\{ -15,4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 2(6x^2-11x-10) $, is \begin{array}{l}\require{cancel} 2(6x^2-15x+4x-10) \\\\= 2[(6x^2-15x)+(4x-10)] \\\\= 2[3x(2x-5)+2(2x-5)] \\\\= 2[(2x-5)(3x+2)] \\\\= 2(2x-5)(3x+2) .\end{array}
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