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 - Page 326: 73



Work Step by Step

Factoring the negative $GCF= -2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -24a^2+34ab-12b^2 \\\\= -2(12a^2-17ab+6b^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -2(12a^2-17ab+6b^2) \end{array} has $ac= 12(6)=72 $ and $b= -17 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,-8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2(12a^2-9ab-8ab+6b^2) .\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} -2[(12a^2-9ab)-(8ab-6b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2[3a(4a-3b)-2b(4a-3b)] .\end{array} Factoring the $GCF= (4a-3b) $ of the entire expression above results to \begin{array}{l}\require{cancel} -2[(4a-3b)(3a-2b)] \\\\= -2(4a-3b)(3a-2b) .\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.