Negative variance of difference of binomial random variables

Question

Let $A = \text{binom}(n, p), B = n - A, D = A - B$
Actually, B is a complimentary binomial random variable of A, so it might as well be written as $B = \text{binom}(n, 1-p)$.

I calculated $E(D) = 2np - n$, but using the formula $V(D) = E(D^2) - E(D)^2$, I am getting $V(D) = -4np^2$ which is negative. How is this possible?

I think the answer should be $4np(1-p)$ because $D = 2A - n \implies V(D) = 4 V(A) + V(n) = 4np(1-p) + 0 = 4np(1-p)$. But I can't get it by algebra.

Answer 1

Yes. The variance will be $4np(1-p)$. As it is mentioned in Andre Nicolas comment. Hence it is not negative. Always positive.

Variance is $\ge 0$.