## Intermediate Algebra (6th Edition)

$q^2(4-p)(16+4p+p^2)$
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}