Resolve each of the following quadratic trinomial into factor:

The given expression is $14 x^{2}+11 x y-15 y^{2} .           \quad$ (Coefficient of $x^{2}=14$, coefficient of $x=11 y$ and constant term $=-15 \mathrm{y}^{2}$ )

Now, we will split the coefficient of $x$ into two parts such that their sum is $11 y$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $14 \times\left(-15 \mathrm{y}^{2}\right)=-210 \mathrm{y}^{2}$.

Now,

$21 y+(-10 y)=11 y$

and

$21 \mathrm{y} \times(-10 \mathrm{y})=-210 \mathrm{y}^{2}$

Replacing the middle term $-11$ xy by $-10 x y+21 x y$, we get:

$14 \mathrm{x}^{2}+11 \mathrm{xy}-15 \mathrm{y}^{2}=14 \mathrm{x}^{2}-10 \mathrm{xy}+21 \mathrm{xy}-15 \mathrm{y}^{2}$

$=\left(14 \mathrm{x}^{2}-10 \mathrm{xy}\right)+\left(21 \mathrm{xy}-15 \mathrm{y}^{2}\right)$

$=2 \mathrm{x}(7 \mathrm{x}-5 \mathrm{y})+3 \mathrm{y}(7 \mathrm{x}-5 \mathrm{y})$

$=(2 \mathrm{x}+3 \mathrm{y})(7 \mathrm{x}-5 \mathrm{y})$