The given zeros are -5, -5, 1 + 2i.
Because complex zeros occur in conjugate pairs, the zeros are
-5, -5, 1 + 2i, 1 - 2i.
The polynomial is
f(x) = (x+5)²[x- (1+2i)][x -(1-2i)]
Note that
[x - (1+2i)][x - (1-2i)]
= x² - x(1 - 2i) - x(1+2i) + (1 - 4i²)
= x² - 2x +5
Therefore
f(x) = (x+5)²(x²-2x+5)
Answer: f(x) = (x+5)²(x²-2x+5)
