Find the zeros of the quadratic polynomial

We have:

$f(x)=4 x^{2}-4 x-3$

$=4 x^{2}-(6 x-2 x)-3$

$=4 x^{2}-6 x+2 x-3$

$=2 x(2 x-3)+1(2 x-3)$

$=(2 x+1)(2 x-3)$

$\therefore f(x)=0=>(2 x+1)(2 x-3)=0$

$=>2 x+1=0$ or $2 x-3=0$

$=>x=\frac{-1}{2}$ or $x=\frac{3}{2}$

So, the zeros of $f(x)$ are $\frac{-1}{2}$ and $\frac{3}{2}$.

Sum of the zeros $=\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{-1+3}{2}=\frac{2}{2}=1=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of the zeros $=\left(\frac{-1}{2}\right) \times \frac{3}{2}=\frac{-3}{4}=\frac{\text { constant term }}{\text { (coefficient of } x^{2} \text { ) }}$