Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$.
=>$(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$
$\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$
$\Rightarrow f(x)=x^{2}-\frac{5}{2} x+1$
$\Rightarrow f(x)=2 x^{2}-5 x+2$
Hence, the required polynomial is $f(x)=2 x^{2}-5 x+2$
$\therefore f(x)=0 \Rightarrow 2 x^{2}-5 x+2=0$
$$
\begin{aligned}
&\Rightarrow 2 \mathrm{x}^{2}-(4 \mathrm{x}+\mathrm{x})+2=0 \\
&\Rightarrow 2 \mathrm{x}^{2}-4 \mathrm{x}-\mathrm{x}+2=0 \\
&\Rightarrow 2 \mathrm{x}(\mathrm{x}-2)-1(\mathrm{x}-2)=0 \\
&\Rightarrow(2 \mathrm{x}-1)(\mathrm{x}-2)=0 \\
&\Rightarrow(2 \mathrm{x}-1)=0 \text { or }(\mathrm{x}-2)=0
\end{aligned}
$$
⇒x=12 or x=2
\Rightarrow \mathrm{x}=\frac{1}{2} \text { or } \mathrm{x}=2
Therefore, the zeros of $f(x)$ are $\frac{1}{2}$ and 2 .