Obtain the zeroes of the polynomial $\displaystyle 7 x^{2}+18 x-9$. Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.

\[\begin{aligned} & 7 x^{2}+18 x-9 \\ = & (7 x-3)(x+3) \\ \therefore & \text { Zeroes are }-3, \frac{3}{7} \end{aligned} \] New zeroes are \(\displaystyle -6, \frac{6}{7}\) Sum of new zeroes \(\displaystyle =(-6)+\frac{6}{7}=-\frac{36}{7}\) Product of new zeroes \(\displaystyle =(-6) \times \frac{6}{7}=-\frac{36}{7}\) ∴ Required polynomial is \(\displaystyle x^{2}+\frac{36}{7} x-\frac{36}{7}\) or \(\displaystyle 7 x^{2}+36 x-36\)