## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 37

#### Answer

$4(x+6)(3x-1)$

#### Work Step by Step

Factoring the $GCF= 4 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 12x^2+68x-24 \\\\= 4(3x^2+17x-6) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 4(3x^2+17x-6) \end{array} has $ac= 3(-6)=-18$ and $b= 17 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 18,-1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4(3x^2+18x-1x-6) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 4[(3x^2+18x)-(x+6)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4[3x(x+6)-(x+6)] .\end{array} Factoring the $GCF= (x+6)$ of the entire expression above results to \begin{array}{l}\require{cancel} 4[(x+6)(3x-1)] \\\\= 4(x+6)(3x-1) .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.