If the zeroes of the quadratic polynomial

(d) Let $p\{x)=x^{2}+(a+1) x+b$

Given that, 2 and $-3$ are the zeroes of the quadratic polynomial $p(x)$.

$\therefore \quad p(2)=0$ and $p(-3)=0$

$\Rightarrow \quad 2^{2}+(a+1)(2)+b=0$

$\Rightarrow \quad 4+2 a+2+b=0$

$\Rightarrow \quad 2 a+b=-6 \quad \ldots$ (i)

and $\quad(-3)^{2}+(a+1)(-3)+b=0$

$\Rightarrow \quad 9-3 a-3+b=0$

$\Rightarrow \quad 3 a-b=6 \quad \ldots$ (ii)

On adding Eqs. (i) and (ii), we get

$5 a=0 \Rightarrow a=0$

Put the value of $a$ in Eq. (i), we get

$2 \times 0+b=-6 \Rightarrow b=-6$

required values are $a=0$ and $b=-6$.