Find the zeroes of the polynomial $\displaystyle \mathrm{p}(x)=3 x^{2}-4 x-4$. Hence, write a polynomial whose each of the zeroes is $\displaystyle 2$ more than zeroes of $\displaystyle \mathrm{p}(x)$.

\[p(x)=3 x^{2}-4 x-4 \] Zeroes are \(\displaystyle -\frac{2}{3}\) and $\displaystyle 2$ New zeroes are \(\displaystyle \frac{4}{3}\) and $\displaystyle 4$ Sum of new zeroes \(\displaystyle =\frac{4}{3}+4=\frac{16}{3}\) Product of new zeroes \(\displaystyle =\frac{4}{3} \times 4=\frac{16}{3}\) Required polynomial is \(\displaystyle x^{2}-\frac{16 x}{3}+\frac{16}{3}\) or \(\displaystyle 3 x^{2}-16 x+16\)