Answer
$$\color{blue}{f(x)=x^2-10x+26}$$
Work Step by Step
We are given zeros and asked to write a polynomial function.
If $5-i$ is a zero, then
$n+5-i=0$
$n-i=-5$
$n=-5+i$
So our factor is $\bf{(x-5+i)}$ because when $x= 5-i $, $(x-5+i)=0$
If $5+i$ is a zero, then
$n+5+i=0$
$n+i=-5$
$n=-5-i$
So our factor is $\bf{(x-5-i)}$ because when $x= 5+i $, $(x-5-i)=0$
So our function is:
$f(x)=(x-5-i)(x-5+i)$
$f(x)=x^2-5x+xi-5x+25-5i-xi+5i-i^2$
$f(x)=x^2-10x+25-i^2$
Recall that $i^2=-1$
$f(x)=x^2-10x+25-(-1)$
$$\color{blue}{f(x)=x^2-10x+26}$$
