#### 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}