Factorize each of the following algebraic expression:

Solution:

We have:

$40+3 x-x^{2}$

$\Rightarrow-\left(x^{2}-3 x-40\right)$

To factorise $\left(\mathrm{x}^{2}-3 \mathrm{x}-40\right)$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=-3$ and $\mathrm{pq}=-40$.

Now,

$5+(-8)=-3$

And

$5 \times(-8)=-40$

Splitting the middle term $-3 \mathrm{x}$ in the given quadratic as $5 \mathrm{x}-8 \mathrm{x}$, we get:

$40+3 x-x^{2}=-\left(x^{2}-3 x-40\right)$

$=-\left(x^{2}+5 x-8 x-40\right)$

$=-\left[\left(x^{2}+5 x\right)-(8 x+40)\right]$

$=-[x(x+5)-8(x+5)]$

$=-(x-8)(x+5)$

$=(x+5)(-x+8)$