Answer
$f(x)=x^{5}-2x^{4}+3x^{3}-2x^{2}+2x$
(sample answer, taking $a=1$)
Work Step by Step
Complex zeros come in conjugate pairs (conjugate zeros theorem).
$i$ is a zero $\Rightarrow -i$ is also a zero.
$1+i$ is a zero $\Rightarrow 1-i$ is also a zero.
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
So,
$(x-0)=x$ is a factor of f,
$(x-i)$ is a factor of f,
$(x+i)$ is a factor of f.
$(x-(1+i))=[(x-1)-i]$ is a factor of f,
$(x-(1-i))=[(x-1)+i]$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=ax(x-i) (x+i) [(x-1)-i] [(x-1)+i]$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
$\left\{\begin{array}{ll}
(x-i) (x+i) & =x^{2}-i^{2}\\
& =x^{2}+1\\
& \\
[(x-1)-i] [(x-1)+i] & =(x-1)^{2}-i^{2}\\
& =x^{2}-2x+1-(-1)\\
& =x^{2}-2x+2
\end{array}\right.$
$ f(x)=x(x^{2}+1)(x^{2}-2x+2)\quad$... distribute,
$ f(x)=x^{3}(x^{2}+1)-2x^{2}(x^{2}+1)+2x(x^{2}+1)\quad$... distribute,
$f(x)=x^{5}+x^{3}-2x^{4}-2x^{2}+2x^{3}+2x$
$f(x)=x^{5}-2x^{4}+3x^{3}-2x^{2}+2x$
