$$
\begin{aligned}
\mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}-11 \mathrm{x}+15 \\
&=2 \mathrm{x}^{2}-(6 \mathrm{x}+5 \mathrm{x})+15 \\
&=2 \mathrm{x}^{2}-6 \mathrm{x}-5 \mathrm{x}+15 \\
=& 2 \mathrm{x}(\mathrm{x}-3)-5(\mathrm{x}-3) \\
\quad=(2 \mathrm{x}-5)(\mathrm{x}-3) \\
\therefore \mathrm{f}(\mathrm{x}) &=0 \Rightarrow(2 \mathrm{x}-5)(\mathrm{x}-3)=0 \\
& \Rightarrow 2 \mathrm{x}-5=0 \text { or } \mathrm{x}-3=0 \\
& \Rightarrow \mathrm{x}=\frac{5}{2} \text { or } \mathrm{x}=3
\end{aligned}
$$

So, the zeroes of $f(x)$ are $\frac{5}{2}$ and 3 .

Sum of zeroes $=\frac{5}{2}+3=\frac{5+6}{2}=\frac{11}{2}=\frac{-(\text { coef ficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of zeroes $=\frac{5}{2} \times 3=\frac{-15}{2}=\frac{\text { constant term }}{\left(\text { coef ficient of } x^{2}\right)}$