Answer
$q^2(4-p)(16+4p+p^2)$
Work Step by Step
Factoring the $GCF=
q^2
$ results to $
q^2(64-p^3)
$. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then,
\begin{array}{l}
q^2(64-p^3)
\\=
q^2[4+(-p)][(4)^2-(4)(-p)+(-p)^2)
\\=
q^2(4-p)(16+4p+p^2)
.\end{array}
