## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 304: 70

#### Answer

$x(4x-1)(3x+1)$

#### Work Step by Step

Factoring the $GCF=x$, then the given expression, $12x^3+x^2-x$, is equivalent to \begin{array}{l} x(12x^2+x-1) .\end{array} The two numbers whose product is $ac= 12(-1)=-12$ and whose sum is $b= 1$ are $\{ -3,4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $x(12x^2+x-1)$, is \begin{array}{l}\require{cancel} x(12x^2-3x+4x-1) \\\\= x[(12x^2-3x)+(4x-1)] \\\\= x[3x(4x-1)+(4x-1)] \\\\= x[(4x-1)(3x+1)] \\\\= x(4x-1)(3x+1) .\end{array}

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