Find the zeros of the quadratic polynomial

We have:

$f(x)=5 x^{2}-4-8 x$

$=5 x^{2}-8 x-4$

$=5 x^{2}-(10 x-2 x)-4$

$=5 x^{2}-10 x+2 x-4$

$=5 x(x-2)+2(x-2)$

$=(5 x+2)(x-2)$

$\therefore f(x)=0=>(5 x+2)(x-2)=0$

$=>5 x+2=0$ or $x-2=0$

$=>x=\frac{-2}{5}$ or $x=2$

So, the zeros of $f(x)$ are $\frac{-2}{5}$ and 2 .

Sum of the zeros $=\left(\frac{-2}{5}\right)+2=\frac{-2+10}{5}=\frac{8}{5}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of the zeros $=\left(\frac{-2}{5}\right) \times 2=\frac{-4}{5}=\frac{\text { constant term }}{\text { (coefficient of } x^{2} \text { ) }}$