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: 8

Answer

$3mn(3m+2n)(2m-n)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 18m^3n+3m^2n^2-6mn^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 $ 3mn ,$ 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} 3mn(6m^2+mn-2n^2) .\end{array} In the trinomial expression above, $a= 6 ,b= 1 ,\text{ and } c= -2 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 6(-2)=-12 $ and whose sum is $b$ are $\left\{ 4,-3 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 3mn(6m^2+4mn-3mn-2n^2) .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} 3mn[(6m^2+4mn)-(3mn+2n^2)] \\\\= 3mn[2m(3m+2n)-n(3m+2n)] \\\\= 3mn[(3m+2n)(2m-n)] \\\\= 3mn(3m+2n)(2m-n) .\end{array}
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