Answer
$f(x)=x^{2}-14x+53$
(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,
$(x-(7-2i))$ is a factor of f,
$(x-(7+2i))$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=a(x-(7-2i))(x-(7+2i))$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
Use FOIL to distribute
$f(x)=x^{2}-x(7+2i)-x(7-2i)+(7-2i)(7+2i)$
... distribute, recognize a difference of squares
$f(x)=x^{2}-7x-2ix-7x+2ix+49-4i^{2}$
... the terms with $ix$ cancel, and $i^{2}=-1$....
$f(x)=x^{2}-7x-7x+49-4(-1)$
$f(x)=x^{2}-14x+53$
