Answer
$f(x)=3x^{4}-x^{3}-9x^{2}+159x-52$
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.
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If $(2+3i)$ is a zero, so is $(2-3i).$
$f(x)=a_{n}(x-\displaystyle \frac{1}{3})(x+4)(x-2+3i)(x-2-3i)$
... factor 1/3 out of the first parentheses...
$f(x)=\displaystyle \frac{a_{n}}{3}(3x-1)(x+4)(x-2+3i)(x-2-3i)$
$f(x)=\displaystyle \frac{a_{n}}{3}(3x^{2}+11x-4)(x^{2}-2x-3ix-2x+4+6i+3ix-6i-9i^{2})$
$f(x)=\displaystyle \frac{a_{n}}{3}(3x^{2}+11x-4)(x^{2}-4x+13)$
Using the given $f(1),$we find $\displaystyle \frac{a_{n}}{3}:$
$f(1)=\displaystyle \frac{a_{n}}{3}(3+11-4)(1-4+13)$
$100=\displaystyle \frac{a_{n}}{3}(10)(10)$
$100=\displaystyle \frac{a_{n}}{3}(100)$
$\displaystyle \frac{a_{n}}{3}=1$
$f(x)=1\cdot(3x^{2}+11x-4)(x^{2}-4x+13)$
$f(x)=3x^{4}-12x^{3}+39x^{2}+11x^{3}-44x^{2}+143x-4x^{2}+16x-52$
$f(x)=3x^{4}-x^{3}-9x^{2}+159x-52$
