Let P be the polynomial you're talking about. I assume P != 0 or it would be trivial.
P has real coefficients, so P(alpha) = 0 => P(baralpha) = 0. It means that there is another root for P, bar(2-i) = 2 + i, hence this form for P :
P(X) = a(X+3)^(a_1)*(X-4)^(a_2)*(X - 2 + i)^(a_3)*(X-2-i)^(a_4)*Q(X) with a_j in NN, Q in RR[X] and a in RR because we want P to have real coefficients.
We want the degree of P to be as small as possible. If R(X) = a(X+3)^(a_1)(X-4)^(a_2)(X - 2 + i)^(a_3)(X-2-i)^(a_4) then deg(P) = deg(R) + deg(Q) = sum (a_j+1) + deg(Q). Q != 0 so deg(Q) >= 0. If we want P to have the smallest degree possible, then deg(Q) = 0 (Q is just a real number q), hence deg(P) = deg(R) and here we can even say that P = R. deg(P) will be as small as possible if each a_j = 0. So deg(P) = 4.
So for now, P(X) = a(X+3)(X-4)(X - 2 + i)(X-2-i)q. Let's develop that.
P(X) = aq(X^2 - X - 12)(X^2-4X+5) in RR[X]. So this expression is the best P we can find with those conditions!
