Answer
$\color{blue}{(3m+n+1)(3m-n-1)}$
Work Step by Step
Group the last three terms together to obtain:
$9m^2-n^2-2n-1=9m^2-(n^2+2n+1)$
The trinomial is a perfect square trinomial in the form $a^2+2ab+b^2$ with $a=n$ and $b=1$.
RECALL:
$a^2-2ab+b^2=(a-b)^2$
Use the formula above to obtain:
$=9m^2-(n+1)^2
\\=(3m)^2-(n+1)^2$
The polynomial above is a difference of two squares.
Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=3m$ and $b=n+1$ to obtain:
$=[3m+(n+1)][3m-(n+1)]
\\=\color{blue}{(3m+n+1)(3m-n-1)}$
