Answer
$f(x)=x^5-5x^4+11x^3-13x^2+8x-2.$
Work Step by Step
The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa.
Therefore the zeros of the function are: $1\pm i$ and $1$ with multiplicity $3$.
Hence the function could e.g. be: $(x-(1+i))(x-(1-i))(x-1)(x-1)(x-1)\\=(x-1-i)(x-1+i)(x-1)^3\\=((x-1)^2-i^2)(x^3 - 3 x^2 + 3 x - 1)\\=(x^2-2x+1+1)(x^3 - 3 x^2 + 3 x - 1)\\=(x^2-2x+2)(x^3 - 3 x^2 + 3 x - 1)\\=x^5-5x^4+11x^3-13x^2+8x-2.$
