Answer
$f(x)=x^{4}-9x^{3}+21x^{2}+21x-130$
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 $(3+2i)$ is a zero, so is $(3-2i).$
$f(x)=a_{n}(x+2)(x-5)(x-3+2i)(x-3-2i)$
$f(x)=a_{n}(x^{2}-3x-10)(x^{2}-3x-2ix-3x+9+6i+2ix-6i-4i^{2})$
$f(x)=a_{n}(x^{2}-3x-10)(x^{2}-6x+13)$
Using the given $f(1),$we find $a_{n}:$
$f(1)=a_{n}(1-3(-1)-10)(1-6(-1)+13)$
$-96=a_{n}(-6)(16)$
$-96=a_{n}(-96)$
$a_{n}=1$
$f(x)=1(x^{2}-3x-10)(x^{2}-6x+13)$
$f(x)=x^{4}-6x+13x^{2}-3x^{3}+18x^{2}-39x-10x^{2}+60x-130$
$f(x)=x^{4}-9x^{3}+21x^{2}+21x-130$
