jmc

We have $36-9x^2=6^2-(3x{)}^2=(6-3x)(6+3x)$. We can factor a 3 out of each of $6-3x$ and $6+3x$ to give $3\cdot (2-x)\cdot 3\cdot (2+x)=\boxed{9(2-x)(2+x)}$. (We could also have factored out a 9 at the beginning: $36-9x^2=9(4-x^2)=9(2-x)(2+x)$.)