Answer
$$\color{blue}{ f(x)= x^5-4x^4+6x^3-4x^2+5x }$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $ 0 $ is a zero, then
$n +0 =0$
$n= 0 $
So our factor is $\bf{(x )}$ because when $x= 0 $, $(x +0 )=0$
If $ -i $ is a zero, then
$n -i =0$
$n= i $
So our factor is $\bf{(x +i )}$ because when $x= -i $, $(x +i )=0$
Since $-i$ is a complex number, its conjugate, $i$ is also a zero
$n +i =0$
$n= -i $
So our factor is $\bf{(x -i )}$ because when $x= i $, $(x -i )=0$
If $ 2+i $ is a zero, then
$n + 2+i =0$
$n +i = -2 $
$n= -2-i $
So our factor is $\bf{(x -2-i )}$
because when $x= 2+i $, $(x -2-i )=0$
Since $ 2+i $ is a complex number, its conjugate, $ 2-i $ is also a zero
$n +2-i =0$
$n-i= -2 $
$n= -2+i $
So our factor is $\bf{(x -2+i )}$
because when $x= 2-i $, $(x -2+i )=0$
So our function is:
$f(x)=(x)(x +i)(x -i )(x -2-i ) (x -2+i) $
$f(x)=(x)(x +i)(x -i )(x^2-4x+4-i^2) $
Recall that $i^2=-1$
$f(x)=(x)(x +i)(x -i )(x^2-4x+4-(-1)) $
$f(x)=(x)(x +i)(x -i )(x^2-4x+5)$
$f(x)=(x)(x^2-ix+ix-i^2)(x^2-4x+5)$
$f(x)=(x)(x^2-(-1))(x^2-4x+5)$
$f(x)=(x)(x^2+1)(x^2-4x+5)$
$f(x)=(x^3+x)(x^2-4x+5)$
$f(x)=x^5-4x^4+5x^3+x^3-4x^2+5x$
$$\color{blue}{ f(x)= x^5-4x^4+6x^3-4x^2+5x }$$
