Answer
$(q+3-p)(q+3+p)$
Work Step by Step
The first $3$ terms of the given expression, $
q^2+6q+9-p^2
,$ is a perfect square trinomial. Hence, the factored form is
\begin{array}{l}\require{cancel}
(q^2+6q+9)-p^2
\\\\=
(q+3)^2-p^2
.\end{array}
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then the factored form of the expression, $
(q+3)^2-p^2
,$ is
\begin{array}{l}\require{cancel}
[(q+3)-p][(q+3)+p]
\\\\=
(q+3-p)(q+3+p)
.\end{array}
