Find the zeroes of the polynomial $\displaystyle \mathrm{q}(x)=8 x^{2}-2 x-3$. Hence, find a polynomial whose zeroes are $\displaystyle 2$ less than the zeroes of $\displaystyle \mathrm{q}(x)$.

\[\begin{aligned} & p(x)=8 x^{2}-2 x-3 \\ & \text { Zeroes are }-\frac{1}{2} \text { and } \frac{3}{4} \\ & \text { New zeroes are }-\frac{5}{2} \text { and }-\frac{5}{4} \\ & \text { Sum of new zeroes }=\frac{-5}{2}+\frac{-5}{4}=\frac{-15}{4} \\ & \text { Product of new zeroes }=\left(-\frac{5}{2}\right) \times\left(-\frac{5}{4}\right)=\frac{25}{8} \\ & \text { Required polynomial is } x^{2}+\frac{15}{4} x+\frac{25}{8} \text { or } 8 x^{2}+30 x+25 \end{aligned} \]