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: 41

Answer

$4pq(3p+5q)(2p+q)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 24p^3q+52p^2q^2+20pq^3 ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 4pq ,$ 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} 4pq(6p^2+13pq+5q^2) .\end{array} In the trinomial expression above, $a= 6 ,b= 13 ,\text{ and } c= 5 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 6(5)=30 $ and whose sum is $b$ are $\left\{ 10,3 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 4pq(6p^2+10pq+3pq+5q^2) .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} 4pq[(6p^2+10pq)+(3pq+5q^2)] \\\\= 4pq[2p(3p+5q)+q(3p+5q)] \\\\= 4pq[(3p+5q)(2p+q)] \\\\= 4pq(3p+5q)(2p+q) .\end{array}
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