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: 5

Answer

$3pq(a+6b)(a-5b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 3a^2pq+3abpq-90b^2pq ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $x^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 3pq ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 3pq(a^2+ab-30b^2) .\end{array} In the trinomial expression above, $a= 1 ,b= 1 ,\text{ and } c= -30 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ 6,-5 \right\}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} 3pq(a+6b)(a-5b) .\end{array}
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