Resolve each of the following quadratic trinomial into factor:

The given expression is $6 \mathrm{x}^{2}-13 \mathrm{xy}+2 \mathrm{y}^{2}$.          (Coefficient of $\mathrm{x}^{2}=6$, coefficient of $\mathrm{x}=-13 \mathrm{y}$ and constant term $=2 y^{2}$ )

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-13 \mathrm{y}$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $6 \times\left(2 \mathrm{y}^{2}\right)=12 \mathrm{y}^{2}$.

Now,

$(-12 y)+(-y)=-13 y$

and

$(-12 y) \times(-y)=12 y^{2}$

Replacing the middle term $-13 \mathrm{xy}$ by $-12 \mathrm{xy}-\mathrm{xy}$, we get:

$6 x^{2}-13 x y+2 y^{2}=6 x^{2}-12 x y-x y+2 y^{2}$

$=\left(6 \mathrm{x}^{2}-12 \mathrm{xy}\right)-\left(\mathrm{xy}-2 \mathrm{y}^{2}\right)$

$=6 \mathrm{x}(\mathrm{x}-2 \mathrm{y})-\mathrm{y}(\mathrm{x}-2 \mathrm{y})$

$=(6 \mathrm{x}-\mathrm{y})(\mathrm{x}-2 \mathrm{y})$