Thanks Alan,
Hi Jurimagic,
Perhaps you have not looked at what Alan has done with a fresh eye.
This is ultimately a difference of 2 squares factorization.
Incidentally, I know, and expect Alan knows, that you are always grateful for the help we give you.
That is why, if I am on and I see your question, I will always answer quickly. 
\(p^2-9s^2+6pq+9q^2\\ =p^2+6pq+9q^2-9s^2\\ =(p^2+6pq+9q^2) - (9s^2)\\ =(p+3q)(p+3q) - (9s^2)\\ =(p+3q)^2 - (9s^2)\\ \text{This is the difference of 2 squares}\\ =(p+3q\;\textcolor{red}{\textbf{-}}\;3s)(p+3q\;\textcolor{red}{\textbf{+}}3s)\)
LaTex Coding:
p^2-9s^2+6pq+9q^2\\
=p^2+6pq+9q^2-9s^2\\
=(p^2+6pq+9q^2) - (9s^2)\\
=(p+3q)(p+3q) - (9s^2)\\
=(p+3q)^2 - (9s^2)\\
\text{This is the difference of 2 squares}\\
=(p+3q\;\textcolor{red}{\textbf{-}}\;3s)(p+3q\;\textcolor{red}{\textbf{+}}3s)