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

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

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

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

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

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

$\Rightarrow {\mathrm{x}}^2–(6\mathrm{x}+2\mathrm{x})+12=0$
\Rightarrow \mathrm{x}^{2}-(6 \mathrm{x}+2 \mathrm{x})+12=0
$\Rightarrow x^2–6x–2x+12=0$
\Rightarrow x^{2}-6 x-2 x+12=0
$\Rightarrow x(x–6)–2(x–6)=0$
\Rightarrow x(x-6)-2(x-6)=0
$\Rightarrow (x–2)(x–6)=0$
\Rightarrow(x-2)(x-6)=0
$\Rightarrow (x–2)=0\text{ or }(x–6)=0$
\Rightarrow(x-2)=0 \text { or }(x-6)=0

 

$\Rightarrow \mathrm{x}=2\text{ or }\mathrm{x}=6$
\Rightarrow \mathrm{x}=2 \text { or } \mathrm{x}=6

So, the zeroes of $\mathrm{f}(\mathrm{x})$ are 2 and 6 .