Resolve each of the following quadratic trinomial into factor:

The given expression is $3+23 y-8 y^{2} .                    \quad$ (Coefficient of $y^{2}=-8$, coefficient of $y=23$ and constant term $=3)$

We will split the coefficient of y into two parts such that their sum is 23 and their product equals the product of the coefficient of $\mathrm{y}^{2}$ and the constant term, i.e., $(-8) \times 3=-24$.

Now,

$(-1)+24=23$

and

$(-1) \times 24=-24$

Replacing the middle term $23 y$ by $-y+24 y$, we have :

$3+23 y-8 y^{2}$

$=-8 \mathrm{y}^{2}+23 \mathrm{y}+3$

$=-8 \mathrm{y}^{2}-\mathrm{y}+24 \mathrm{y}+3$

$=\left(-8 \mathrm{y}^{2}-\mathrm{y}\right)+(24 \mathrm{y}+3)$

$=-\mathrm{y}(8 \mathrm{y}+1)+3(8 \mathrm{y}+1)$

$=(3-\mathrm{y})(8 \mathrm{y}+1)$