Answer
$x^4-4x^3+10x^2-12x+5$
Work Step by Step
The Complex Conjugate zeroes Theorem states that the conjugate of $i$ is also a zero of the polynomial $P(x)$.
Here, we have three zeroes of the polynomial $P(x)$ of degree $4$.
The factorization of $P(x)$ is given as follows:
$P(x)=(x-1)^2[x-(1-2i)][x-(1+2i)]$
Apply the difference square formula.
$P(x)=(x^2-2x+1)[(x-1)^2-(2i)^2]=(x^2-2x+1)(x^2-2x+1-4i^2)=(x^2-2x+1)(x^2-2x+5)$
Hence, $P(x)=x^4-4x^3+10x^2-12x+5$
