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!