Factorize:

We have:

$9(2 a-b)^{2}-4(2 a-b)-13$

Let :

$(2 a-b)=p$

Thus, the given expression becomes

$9 p^{2}-4 p-13$

Now, we must split $(-4)$ into two numbers such that their sum is $(-4)$ and their product is $(-117)$.

Clearly, $-13+9=-4$ and $-13 \times 9=-117$.

$\therefore 9 p^{2}-4 p-13=9 p^{2}+9 p-13 p-13$

$=9 p(p+1)-13(p+1)$

$=(p+1)(9 p-13)$

Putting $p=(2 a-b)$, we get:

$9(2 a-b)^{2}-4(2 a-b)-13=[(2 a-b)+1][9(2 a-b)-13]$

$=(2 a-b+1)[18 a-9 b-13]$