Resolve each of the following quadratic trinomial into factor:

The given expression is $6 \mathrm{x}^{2}-5 \mathrm{xy}-6 \mathrm{y}^{2}$.         (Coefficient of $\mathrm{x}^{2}=6$, coefficient of $\mathrm{x}=-5 \mathrm{y}$ and constant term $=-6 \mathrm{y}^{2}$ )

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-5 \mathrm{y}$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $6 \times\left(-6 \mathrm{y}^{2}\right)=-36 \mathrm{y}^{2} .$

Now,

$(-9 y)+4 y=-5 y$

and

$(-9 y) \times 4 y=-36 y^{2}$

Replacing the middle term $-5 x y$ by $-9 x y+4 x y$, we get:

$6 x^{2}-5 x y-6 y^{2}=6 x^{2}-9 x y+4 x y-6 y^{2}$

$=\left(6 x^{2}-9 x y\right)+\left(4 x y-6 y^{2}\right)$

$=3 x(2 x-3 y)+2 y(2 x-3 y)$

$=(3 x+2 y)(2 x-3 y)$