Answer
$f(x)=x^{5}-4x^{4}+6x^{3}-4x^{2}+5x$
(sample answer with $a=1$)
Work Step by Step
Complex zeros come in conjugate pairs (conjugate zeros theorem).
$-i$ is a zero $\Rightarrow +i$ is also a zero.
$2+i$ is a zero $\Rightarrow 2-i$ 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.
$(x-(2+i))=[(x-2)-i]$ is a factor of f,
$(x-(2-i))=[(x-2)+i]$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=ax(x-i) (x+i) [(x-2)-i] [(x-2)+i]$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
Some of the products are differences of squares:
$\left\{\begin{array}{ll}
(x-i) (x+i) & =x^{2}-i^{2}\\
& =x^{2}-(-1)\\
& =x^{2}+1\\
& \\
[(x-2)-i] [(x-2)+i] & =(x-1)^{2}-i^{2}\\
& =x^{2}-4x+4-(-1)\\
& =x^{2}-4x+5
\end{array}\right.$
$ f(x)=x(x^{2}+1)(x^{2}-4x+5)\quad$... distribute,
$ f(x)=x^{3}(x^{2}+1)-4x^{2}(x^{2}+1)+5x(x^{2}+1)\quad$... distribute,
$f(x)=x^{5}+x^{3}-4x^{4}-4x^{2}+5x^{3}+5x$
$f(x)=x^{5}-4x^{4}+6x^{3}-4x^{2}+5x$
(sample answer )
