#### Answer

$y^2(2x+1)(x^2-7)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Factor the $GCF$ of the terms. Then, group the terms of the given expression, $
2x^3y^2+x^2y^2-14xy^2-7y^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 $GCF=
y^2
,$, the expression above is equivalent to
\begin{array}{l}\require{cancel}
y^2\left( \dfrac{2x^3y^2}{y^2}+\dfrac{x^2y^2}{y^2}-\dfrac{14xy^2}{y^2}-\dfrac{7y^2}{y^2} \right)
\\\\=
y^2\left( 2x^3+x^2-14x-7 \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}
y^2[(2x^3+x^2)-(14x+7)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
y^2[x^2(2x+1)-7(2x+1)]
.\end{array}
Factoring the $GCF=
(2x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
y^2[(2x+1)(x^2-7)]
\\\\=
y^2(2x+1)(x^2-7)
.\end{array}