Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12.

Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$.

Then $(\alpha+\beta)=8$ and $\alpha \beta=12$

$\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$

$=>f(x)=x^{2}-8 x+12$

Hence, required polynomial $f(x)=x^{2}-8 x+12$

$\therefore f(x)=0=>x^{2}-8 x+12=0$

$=>x^{2}-(6 x+2 x)+12=0$

$=>x^{2}-6 x-2 x+12=0$

$=>x(x-6)-2(x-6)=0$

$=>(x-2)(x-6)=0$

$=>(x-2)=0$ or $(x-6)=0$

$=>x=2$ or $x=6$

So, the zeroes of $f(x)$ are 2 and 6 .