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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 34

Answer

$-4(x+3)(3x-2)$

Work Step by Step

Factoring the $GCF= -4 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -12x^2-28x+24 \\\\= -4(3x^2+7x-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+7x-6) \end{array} has $ac= 3(-6)=-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} -4(3x^2+9x-2x-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+9x)-(2x+6)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -4[3x(x+3)-2(x+3)] .\end{array} Factoring the $GCF= (x+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} -4[(x+3)(3x-2)] \\\\= -4(x+3)(3x-2) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.