Factorize each of the following algebraic expression:

To factorise $\mathrm{x}^{2}-4 \mathrm{x}-21$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=-4$ and $\mathrm{pq}=-21$.

Now,

$3+(-7)=-4$

and

$3 \times(-7)=-21$

Splitting the middle term $-4 \mathrm{x}$ in the given quadratic as $-7 \mathrm{x}+3 \mathrm{x}$, we get:

$\mathrm{x}^{2}-4 \mathrm{x}-21=\mathrm{x}^{2}-7 \mathrm{x}+3 \mathrm{x}-21$

$=\left(\mathrm{x}^{2}-7 \mathrm{x}\right)+(3 \mathrm{x}-21)$

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

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