Resolve each of the following quadratic trinomial into factor:

The given expression is $12 \mathrm{x}^{2}-17 \mathrm{xy}+6 \mathrm{y}^{2} .          \quad$ (Coefficient of $\mathrm{x}^{2}=12$, coefficient of $\mathrm{x}=-17 \mathrm{y}$ and constant term $=6 \mathrm{y}^{2}$ )

We willsplit the coefficient of $x$ into two parts such that their sum is $-17 y$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term i.e., $12 \times 6 \mathrm{y}^{2}=72 \mathrm{y}^{2}$.

Now,

$(-9 y)+(-8 y)=-17 y$

and

$(-9 \mathrm{y}) \times(-8 \mathrm{y})=72 \mathrm{y}^{2}$

Replacing the middle term $-17$ xy by $-9 x y-8 x y$, we get:

$12 \mathrm{x}^{2}-17 \mathrm{xy}+6 \mathrm{y}^{2}=12 \mathrm{x}^{2}-9 \mathrm{xy}-8 \mathrm{xy}+6 \mathrm{y}^{2}$

$=\left(12 \mathrm{x}^{2}-9 \mathrm{xy}\right)-\left(8 \mathrm{xy}-6 \mathrm{y}^{2}\right)$

$=3 \mathrm{x}(4 \mathrm{x}-3 \mathrm{y})-2 \mathrm{y}(4 \mathrm{x}-3 \mathrm{y})$

$=(3 \mathrm{x}-2 \mathrm{y})(4 \mathrm{x}-3 \mathrm{y})$