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: 74

Answer

$5(3a-4b)(a+b)$

Work Step by Step

Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 15a^2-5ab-20b^2 \\\\= 5(3a^2-ab-4b^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(3a^2-ab-4b^2) \end{array} has $ac= 3(-4)=-12 $ and $b= -1 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -4,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(3a^2-4ab+3ab-4b^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} 5[(3a^2-4ab)+(3ab-4b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[a(3a-4b)+b(3a-4b)] .\end{array} Factoring the $GCF= (3a-4b) $ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(3a-4b)(a+b)] \\\\= 5(3a-4b)(a+b) .\end{array}
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