If α, β, are the zeroes of the polynomial x² – p(x + 1) – c such that (α + 1) (β + 1) = 0 then the value of c will be:

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1

b = -p

c = -p-c

Sum of zeroes = -b/a

= - (-p) / 1

α + β = p …………………………... (i)

Product of Zeroes = c/a

αβ = - (p + c) …………………… (ii)

Now from LHS we have,

(α + 1)(β + 1) = αβ + α + β + 1

From (i) and (ii) we have the values of αβ and (α + β)

(α + 1)(β + 1) = - (p + c) + p + 1

= - p – c + p + 1

= c + 1

Given (α + 1) (β + 1) = 0

c + 1 = 0

c = -1

So the correct answer is B [-1].