Resolve each of the following quadratic trinomial into factor:

The given expression is $2 x^{2}-3 x-2$.                 (Coefficient of $\mathrm{x}^{2}=2$, coefficient of $\mathrm{x}=-3$ and constant term $=-2$ )

We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-3$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $2 \times(-2)=-4$.

Now,

$(-4)+1=-3$

and

$(-4) \times 1=-4$

Replacing the middle term $3 \mathrm{x}$ by $-4 \mathrm{x}+\mathrm{x}$, we have:

$2 x^{2}-3 x-2=2 x^{2}-4 x+x-2$

$=\left(2 \mathrm{x}^{2}-4 \mathrm{x}\right)+(\mathrm{x}-2)$

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

$=(2 \mathrm{x}+1)(\mathrm{x}-2)$