If the zeros of the quadratic polynomial x^{2} + (a + 1)x + b are 2 and - 3,

Then

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.

2 and - 3 are the zeroes of the polynomial p(x) = x^{2} + (a + 1)x + b

i.e. p(2) = 0 and p(- 3) = 0

p(2) = (2)^{2} + (a + 1)(2) + b = 0

= 4 + 2a + 2 + b = 0

= 6 + 2a + b = 0 (1)

P(- 3) = (- 3)^{2} + 9 + (a + 1)(- 3) + b = 0

= 9 - 3a - 3 + b = 0

= 6 - 3a + b = 0 (2)

Equating (1) = (2), as both the equations are equal to zero. Hence both equations are equal to each other.

6 + 2a + b = 6 - 3a + b

= 5a = 0

⇒ a = 0

Putting the value of ‘a’ in (1)

6 + 2(0) + b = 0

⇒ b = - 6

OR

The equation of a quadratic polynomial is given by x^{2} - (sum of the zeroes) x + (product of the zeroes)

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x^{2}

⇒ sum of the zeroes = - (a + 1)

= 2 - 3 = - a - 1

= - 1 + 1 = - a

= - a = 0

⇒ a = 0

Product of the zeroes = constant term ÷ coefficient of x^{2}

⇒ b = product of the zeroes

= 2(- 3)

= 6

6