## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 5 - Polynomials and Factoring - Mid-Chapter Review - Mixed Review - Page 335: 20

#### Answer

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

#### Work Step by Step

Factoring the $GCF= 2n ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 2m^3n-10m^2n-6mn+30n \\\\= 2n(m^3-5m^2-3m+15) .\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} 2n(m^3-5m^2-3m+15) \\\\= 2n[(m^3-5m^2)-(3m-15)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2n[m^2(m-5)-3(m-5)] .\end{array} Factoring the $GCF= (m-5)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2n[(m-5)(m^2-3)] \\\\= 2n(m-5)(m^2-3) .\end{array}

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