Answer
$f(x)=x^{2}-10x+26$
(sample answer, taking $a=1$)
Work Step by Step
Complex zeros come in conjugate pairs (conjugate zeros theorem).
The given zeros are a conjugate pair, so there is no need to search further.
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
So,
$(x-(5+i))$ is a factor of f,
$(x-(5-i))$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=a(x-(5+i))(x-(5-i))$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
Use FOIL to distribute
$f(x)=x^{2}-x(5-i)-x(5+i)+(5+i)(5-i)$
... distribute, recognize a difference of squares
$f(x)=x^{2}-5x+ix-5x-ix+25-i^{2}$
... the terms with $ix$ cancel, and $i^{2}=-1$....
$f(x)=x^{2}-5x-5x+25-(-1)$
$f(x)=x^{2}-10x+26$