Obtain the zeroes of the polynomial $\displaystyle \mathrm{p}(x)=2 x^{2}-5 x-3$. Hence, obtain a polynomial each of whose zeroes is one less than each of the zero of $\displaystyle \mathrm{p}(x)$.

\[\begin{aligned} p(x) & =2 x^{2}-5 x-3 \\ & =(x-3)(2 x+1) \end{aligned} \] ∴ Zeroes are $\displaystyle 3$, \(\displaystyle -\frac{1}{2}\)New zeroes are $\displaystyle 2$, \(\displaystyle -\frac{3}{2}\)Sum of new zeroes \(\displaystyle =2+\left(-\frac{3}{2}\right)=\frac{1}{2}\)Product of new zeroes \(\displaystyle =2 \times\left(-\frac{3}{2}\right)=-3\)∴ Required polynomial is \(\displaystyle x^{2}-\frac{1}{2} x-3\) or \(\displaystyle 2 x^{2}-x-6\)