How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\end{align}
Then I use quadratic formula on numerator to factor it :
$a=2,b=-1,c=-1$
$$=\frac{2(x+2)(x-\frac{5}{2})}{(x-3)(2x+1)}$$
But apparently this can be factored further.  What else can I do?

HINT:
$2x^2-x-1=(x-1)(2x+1)$
$x^2-9=(x+3)(x-3)$
