Resolve each of the following quadratic trinomial into factor:

The given expression is $7 \mathrm{x}^{2}-19 \mathrm{x}-6$.       (Coefficient of $\mathrm{x}^{2}=7$, coefficient of $\mathrm{x}=-19$ and constant term $=$ $-6)$

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-19$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $7 \times(-6)=-42$.

Now,

$(-21)+2=-19$

and

$(-21) \times 2=-42$

Replacing the middle term $-19 x$ by $-21 x+2 x$, we have :

$7 \mathrm{x}^{2}-19 \mathrm{x}-6=7 \mathrm{x}^{2}-21 \mathrm{x}+2 \mathrm{x}-6$

$=\left(7 \mathrm{x}^{2}-21 \mathrm{x}\right)+(2 \mathrm{x}-6)$

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

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