Answer
$P(x)=4x^{4}-16x^{3}+40x-48x+20$
Work Step by Step
Since $1,1-2i$ are zeros, then so $1+2i$ by the Conjugate Zeros Theorem. This mean that P(x) have the following form:
$P(x)=a(x-1)^{2}[x-(1-2i)][x-(1+2i)]$
$P(x)=a(x-1)^{2}[x-1+2i][x-1-2i]$
$P(x)=a(x-1)^{2}[(x-1)^{2}-4i^{2}]$
$P(x)=a(x^{2}-2x+1)(x^{2}-2x+5)$
$P(x)=a(x^{4}-2x^{3}+5x^{2}-2x^{3}+4x^{2}-10x+x^{2}-2x+5)$
$P(x)=a(x^{4}-4x^{3}+10x-12x+5)$
To make all all coefficients integer, we set $a=4$ and get
$P(x)=4x^{4}-16x^{3}+40x-48x+20$
