Resolve each of the following quadratic trinomial into factor:

The given expression is $7 \mathrm{x}-6 \mathrm{x}^{2}+20 .         \quad$ (Coefficient of $\mathrm{x}^{2}=-6$, coefficient of $\mathrm{x}=7$ and constant term $=20$ )

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is 7 and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i. e., $(-6) \times 20=-120$.

Now,

$15+(-8)=7$

and

$15 \times(-8)=-120$

Replacing the middle term $7 \mathrm{x}$ by $15 \mathrm{x}-8 \mathrm{x}$, we get:

$7 \mathrm{x}-6 \mathrm{x}^{2}+20$

$=-6 \mathrm{x}^{2}+7 \mathrm{x}+20$

$=-6 \mathrm{x}^{2}+15 \mathrm{x}-8 \mathrm{x}+20$

$=\left(-6 \mathrm{x}^{2}+15 \mathrm{x}\right)+(-8 \mathrm{x}+20)$

$=3 \mathrm{x}(-2 \mathrm{x}+5)+4(-2 \mathrm{x}+5)$

$=(3 \mathrm{x}+4)(-2 \mathrm{x}+5)$