## 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: 40

#### Answer

$3(2x+3)(3x-1)$

#### Work Step by Step

Rearranging the terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -9+18x^2+21x \\\\= 18x^2+21x-9 .\end{array} Factoring the $GCF= 3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 3(6x^2+7x-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3(6x^2+7x-3) \end{array} has $ac= 6(-3)=-18$ and $b= 7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 9,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3(6x^2+9x-2x-3) .\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} 3[(6x^2+9x)-(2x+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[3x(2x+3)-(2x+3)] .\end{array} Factoring the $GCF= (2x+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(2x+3)(3x-1)] \\\\= 3(2x+3)(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.