Answer
$ x^{6}-12x^{5}+55x^{4}-120x^{3}+139x^{2}-108x+85$
Work Step by Step
If $u$ and $v $ are zeros of $x^{2}+bx+c ,$
then $(x-u)(x-v)=x^{2}-(u+v)x+uv$
equals $x^{2}+bx+c ,$
from where we extract: $b=-(u+v),\quad c=uv$
Since $i$ is a zero, its conjugate $-i$ is also a zero,
$(x-i)$ and $(x+i)$ are factors of f.
$(x-i)(x+i)=x^{2}+1$ is a factor of f.
Since $4-i$ is a zero, its conjugate $4+i$ is also a zero.
$b=-(u+v)=-(4-i+4+i)=-8$
$c=(4-i)(4+i)=16+1=17$
A quadratic binomial with these zeros is $x^{2}-8x+17 ,$
and, having the same factors, it is a factor of f.
Since $2+i$ is a zero, its conjugate $2-i$ is also a zero,
$b=-(u+v)=-(2-i+2+i)=-4$
$c=(2-i)(2+i)=4+1=5$
A quadratic binomial with these zeros is $x^{2}-4x+5 ,$
and, having the same factors, it is a factor of f.
So, letting the leading coefficient be 1,
$f(x)=(x^{2}+1)(x^{2}-8x+17)(x^{2}-4x+5)$
$= (x^{4}-8x^{3}+17x^{2}+x^{2}-8x+17 ) (x^{2}-4x+5 )$
$= (x^{4}-8x^{3}+18x^{2}-8x+17 ) (x^{2}-4x+5 )$
$=x^{6}-4x^{5}+5x^{4}-8x^{5}+32x^{4}-40x^{3}+18x^{4}-72x^{3}+90x^{2}-8x^{2}-40x+17x^{2}-68x+85$
$=x^{6}-12x^{5}+55x^{4}-120x^{3}+139x^{2}-108x+85$
