If –2 and 3 are the zeros of the quadratic polynomial x2 + (a + 1) x + b then
It is given in the question that, -2 and 3 are the zeros of x2 + (a + 1) x + b
∴ (- 2)2 + (a + 1) × (- 2) + b = 0
4 – 2a – 2 + b = 0
b – 2a = - 2 (i)
Also, we have
32 + (a + 1) × 3 + b = 0
9 + 3a + 3 + b = 0
b + 3a = - 12 (ii)
Now, by subtracting the equation (i) from (ii) we get:
a = - 2
Also, b = - 2 – 4 = - 6