Answer
$6x^4-12x^3+18x^2-12x+12$
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 four zeroes of the polynomial $P(x)$ of degree $4$.
The factorization of $P(x)$ is given as follows:
$P(x)=6(x-i)(x+i)[x-(1+i)][x-(1-i)]$
Apply the difference square formula.
$P(x)=6(x^2-i^2)[(x-1)^2-i^2]=6(x^2+1)[(x^2-2x+1)+1]=6(x^4-2x^3+2x^2+x^2-2x+2)$
Hence, $P(x)=6(x^4-2x^3+3x^2-2x+2)=6x^4-12x^3+18x^2-12x+12$
