Find the zeros of the quadratic polynomial

We have:

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

It can be written as $8 x^{2}+0 x-4$

$=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\}$

$=4(\sqrt{2} x+1)(\sqrt{2} x-1)$

$\therefore f(x)=0=>(\sqrt{2} x+1)(\sqrt{2} x-1)=0$

$=>\sqrt{2} x+1=0$ or $\sqrt{2} x-1=0$

$\Rightarrow x=\frac{-1}{\sqrt{2}}$ or $x=\frac{1}{\sqrt{2}}$

So, the zeroes of $f(x)$ are $\frac{-1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$

Here the coefficient of $x$ is 0 and the coefficient of $x^{2}$ is $\sqrt{2}$

Sum of the zeroes $=\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{-1+1}{\sqrt{2}}=\frac{0}{\sqrt{2}}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of the zeroes $=\frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{-1 \times 4}{2 \times 4}=\frac{-4}{8}=\frac{\text { constant term }}{\left(\text { coefficient of } x^{2}\right)}$