Answer
$f(x)=x^4-6x^3+14x^2-16x+8$
Work Step by Step
If $1+i$ is a zero of $f$ then the complex conjugate $1-i$ is also a zero.
$f(x)=a(x-2)(x-2)[x-(1+i)][x-(1-i)]$
$f(x)=a(x-2)^2[(x-1)-i][(x-1)+i]$
$f(x)=a(x^2-4x+4)[(x-1)^2-i^2]$
$f(x)=a(x^2-4x+4)(x^2-2x+1+1)$
$f(x)=a(x^2-4x+4)(x^2-2x+2)$
$f(x)=a(x^4-2x^3+2x^2-4x^3+8x^2-8x+4x^2-8x+8)$
We can choose any value for $a$. So, let's make $a=1$
$f(x)=x^4-6x^3+14x^2-16x+8$
