Answer
$f(x)=x^{4}+10x^{2}+9$
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 $(i)$ is a zero, so is $(-i).$
If $(3i)$ is a zero, so is $(-3i).$
$f(x)=a_{n}(x-i)(x+i)(x-3i)(x+3i)$
$f(x)=a_{n}(x^{2}-i^{2})(x^{2}-9i^{2})$
...$ i^{2}=-1$ ...
$f(x)=a_{n}(x^{2}+1)(x^{2}+9)$
Using the given $f(-1),$we find $a_{n}:$
$f(-1)=a_{n}(1+1)(1+9)$
$20=a_{n}(20)$
$a_{n}=1$
$f(x)=1\cdot(x^{2}+1)(x^{2}+9)$
$f(x)=x^{4}+10x^{2}+9$
