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: 40

Answer

$(3k-5q)(4k+q)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 12k^2-17kq-5q^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 12 ,b= -17 ,\text{ and } c= -5 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 12(-5)=-60 $ and whose sum is $b$ are $\left\{ -20,3 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 12k^2-20kq+3kq-5q^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (12k^2-20kq)+(3kq-5q^2) \\\\= 4k(3k-5q)+q(3k-5q) \\\\= (3k-5q)(4k+q) .\end{array}
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