#### Answer

$f(x)=x^{3}-3x^{2}-15x+125$

#### Work Step by Step

The Linear Factorization Theorem... in words:
An nth-degree polynomial can be expressed as the product of a nonzero constant and $n$ linear factors, where each linear factor has a leading coefficient of 1.
Imaginary roots, if they exist, occur in conjugate pairs.
-------------------------
If $4+3i$ is a zero, so is $4-3i.$
$f(x)=a_{n}(x+5)(x-4-3i)(x-4+3i)$
$f(x)=a_{n}(x+5)(x^{2}-4x+3ix-4x+16-12i-3ix+12i-9i^{2})$
...$ i^{2}=-1$ ...
$f(x)=a_{n}(x+5)(x^{2}-8x+25)$
Using the given $f(2),$we find $a_{n}:$
$f(2)=a_{n}(2+5)(4-16+25)$
$91=a_{n}(7)(13)$
$91=a_{n}(91)$
$a_{n}=1$
so
$f(x)=1\cdot(x+5)(x^{2}-8x+25)$
$f(x)=(x^{3}-8x^{2}+25x+5x^{2}-40x+125)$
$f(x)=x^{3}-3x^{2}-15x+125$