Resolve each of the following quadratic trinomial into factor:

The given expression is $28-31 \mathrm{x}-5 \mathrm{x}^{2}$.            (Coefficient of $\mathrm{x}^{2}=-5$, coefficient of $\mathrm{x}=-31$ and constant term $=28)$

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-31$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $(-5) \times(28)=-140$.

Now,

$(-35)+4=-31$

and

$(-35) \times 4=-140$

Replacing the middle term $-31 \mathrm{x}$ by $-35 \mathrm{x}+4 \mathrm{x}$, we have :

$-5 x^{2}-31 x+28=-5 x^{2}-35 x+4 x+28$

$=\left(-5 x^{2}-35 x\right)+(4 x+28)$

$=-5 x(x+7)+4(x+7)$

$=(4-5 x)(x+7)$