## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 51

#### Answer

$2(x-5)(x+4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $2x^2-2x-40 ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $x^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $2 ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 2(x^2-x-20) .\end{array} In the trinomial expression above, $b= -1 ,\text{ and } c= -20 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -5,4 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} 2(x-5)(x+4) .\end{array}

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