Find a polynomial function f(x), with real coefficients, that has 1 and 3+2i as zeros, and such that f(-1)=2 (Multiply out and simplify your answer)

Answer 1

f(x) = a(x-1)(x-(3+2i))(x-(3-2i))
= a(x-1)(x^2-6x+13)
= a(x^3-7x^2+19x-13

f(-1) = a(-1-7-19-13) = 40a = 2
so, a = 1/20