Answer
$x^4-2x^3+6x^2-2x+5$
Work Step by Step
Let us consider that $a$ is a zero of a function with multiplicity $b$. Then this factor of the function can be expressed as: $(x-a)^b$.
The Conjugate Pairs Theorem states that when a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. This means that, when $(p +i \ q)$ is a zero of a polynomial function with a real number of the coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We are given that the zeros of the function are: $1\pm2i$ and $\pm i$.
Therefore, we can write the equation of the function as:
$f(x)=[x-(1+2i)][x-(1-2i)](x-i)(x-(-i))\\=(x-1-2i)(x-1+2i)(x-i)(x+i)\\=x^4-2x^3+6x^2-2x+5$