Answer
$f(x)=x^4-2x^3+6x^2-2x+5$.
Work Step by Step
If $a$ is a zero of a function with multiplicity $b$ then $(x-a)^b$ is a “multiplier” of the function.
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: $1\pm2i$ and $\pm i$, hence the function could e.g. be: $a(x-(1+2i)(x-(1-2i)(x-i)(x-(-i))\\=a(x-1-2i)(x-1+2i)(x-i)(x+i)\\=a(x^4-2x^3+6x^2-2x+5).$
If $a=1$, the function is: $f(x)=x^4-2x^3+6x^2-2x+5$.
(The degree is $4$ which is good).