Given, the quadratic polynomial is $x^2 + (a + 1)x + b.$The zeros of the polynomial are 2 and -3.We have to find the value of a and b.We know that if α and β are the zeroes of a polynomial $ax^2 + bx + c,$ thenSum of the roots is $\alpha + \beta = \frac{-b}{a}$Product of the roots is $\alpha \beta = \frac{c}{a}$Here, $\alpha = 2$ and $\beta = -3$Coefficient b = (a + 1)Coefficient a = 1Coefficient c = bSum of the roots = -b/a = -(a+1)/1 = -1 - a$\alpha + \beta = 2-3 = -1$So, $-1 = -1 - a$$-1 + a = -1$$a = -1 + 1$a = 0Product of the roots = c/a = b/(a+1)= b/(0+1) = b$\alpha \beta = (2)(-3) = -6$So, b = -6Therefore, the values of a and b are 0 and -6.