If $\displaystyle \alpha$ and $\displaystyle \beta$ are the zeroes of the polynomial $\displaystyle a x^{2}-x+c$. Obtain a polynomial whose zeroes are $\displaystyle \alpha-3$ and $\displaystyle \beta-3$.

\[\begin{aligned} & \alpha+\beta=\frac{1}{a}, \alpha \beta=\frac{c}{a} \\ & \begin{aligned} \text { Sum of zeroes of required polynomial } & =\alpha+\beta-6 \\ & =\frac{1}{a}-6 \text { or } \frac{1-6 a}{a} \end{aligned} \\ & \begin{aligned} \text { Product of zeroes of required polynomial } & =\alpha \beta-3(\alpha+\beta)+9 \\ & =\frac{c}{a}-\frac{3}{a}+9 \end{aligned} \end{aligned} \] ∴ required polynomial is \(\displaystyle x^{2}-\left(\frac{1-6 a}{a}\right) x+\frac{c-3+9 a}{a}\) or \(\displaystyle a x^{2}-(1-6 a) x+(c-3+9 a)\)