## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises: 58

#### Answer

$6(a+4)(3a-8)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $18a^2+24a-192 ,$ factor first the $GCF.$ Then find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Factoring the $GCF= 6 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6(3a^2+4a-32) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $3(-32)=-96$ and the value of $b$ is $4 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 12,-8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6(3a^2+12a-8a-32) .\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} 6[(3a^2+12a)-(8a+32)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 6[3a(a+4)-8(a+4)] .\end{array} Factoring the $GCF= (a+4)$ of the entire expression above results to \begin{array}{l}\require{cancel} 6[(a+4)(3a-8)] \\\\= 6(a+4)(3a-8) .\end{array}

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