Find the polynomial function f with real coefficients that has given degree, zeros, and solution point.

Question

Degree: 3
Zeros: -2,1- root of 2i
Solution point: f(-1)=-54

Answer 1

since complex solutions appear in conjugate pairs, the third solution must be 1 + √2 i
so the function has the form
f(x) = a(x+2)(x - 1 + √2 i)(x - 1 - √2 i)
= a(x-2)(x^2 - 2x + 3)
since f(-1) = a(-3)(1 + 2 + 3) = -18a
so -18a = -54
a = 3

f(x) = 3(x+2)(x^2 - 2x + 3)