Factorize:

We have:

$2(x+y)^{2}-9(x+y)-5$

Let :

$(x+y)=u$

Thus, the given expression becomes

$2 u^{2}-9 u-5$

$2 u^{2}-9 u-5$

Now, we have to split $(-9)$ into two numbers such that their sum is $(-9)$ and their product is $(-10)$.

Clearly, $-10+1=-9$ and $-10 \times 1=-10$

$\therefore 2 u^{2}-9 u-5=2 u^{2}-10 u+u-5$

$=2 u(u-5)+1(u-5)$

$=(u-5)(2 u+1)$

Putting $u=(x+y)$, we get:

$2(x+y)^{2}-9(x+y)-5=(x+y-5)[2(x+y)+1]$

$=(x+y-5)(2 x+2 y+1)$