Let p(x)=x2+(a+1)x+b
Given that, 2 and -3 are the zeroes of the quadratic polynomial p (x)
∴ p(2)=0 and p(−3)=0,
Solving for p(2)
⇒ 22+(a+1)(2)+b=0
⇒ 4+2a+2+b=0
⇒ 2a+b=−6 ....(i)
Solving for p(−3),
(−3)2+(a+1)(−3)+b=0
⇒ 9–3a–3+b=0
⇒ 3a–b=6 ....(ii)
On adding Eqs.(i) and (ii), we get
5a=0 ⇒ a=0
Put the value of a in Eq. (i), we get
x×0+b=−6⇒ b=−6
so, the required values are a=0 and b=−6