Answer
$9q^2+30q+25-p^2$
Work Step by Step
Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the product of the given expression, $
[(3q+5)-p][(3q+5)+p]
,$ is
\begin{array}{l}\require{cancel}
(3q+5)^2-(p)^2
\\\\=
(3q+5)^2-p^2
.\end{array}
Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the expression, $
(3q+5)^2-p^2
,$ is equivalent to
\begin{array}{l}\require{cancel}
(3q)+2(3q)(5)+(5)^2-p^2
\\\\=
9q^2+30q+25-p^2
.\end{array}
