Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 59

Answer

$(7a-4b)(3a+b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 21a^2-5ab-4b^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 21 ,b= -5 ,\text{ and } c= -4 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 21(-4)=-84 $ and whose sum is $b$ are $\left\{ -12,7 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 21a^2-12ab+7ab-4b^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (21a^2-12ab)+(7ab-4b^2) \\\\= 3a(7a-4b)+b(7a-4b) \\\\= (7a-4b)(3a+b) .\end{array}
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