Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises: 72

Answer

$m^2(n^2+5)(3n-2)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor the $GCF$ of the terms. Then, group the terms of the given expression, $ 3m^2n^3+15m^2n-2m^2n^2-10m^2 ,$ such that the factored form of the groupings will result to a factor common to the entire expression. Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression. $\bf{\text{Solution Details:}}$ Using the $GCF= m^2 ,$, the expression above is equivalent to \begin{array}{l}\require{cancel} m^2\left( \dfrac{3m^2n^3}{m^2}+\dfrac{15m^2n}{m^2}-\dfrac{2m^2n^2}{m^2}-\dfrac{10m^2}{m^2} \right) \\\\= m^2\left( 3n^3+15n-2n^2-10 \right) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} m^2[(3n^3+15n)-(2n^2+10)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} m^2[3n(n^2+5)-2(n^2+5)] .\end{array} Factoring the $GCF= (n^2+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} m^2[(n^2+5)(3n-2)] \\\\= m^2(n^2+5)(3n-2) .\end{array}
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