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

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