Answer
The complex zeros are: $2$ with multiplicity $2, i \sqrt 5$, and $-i \sqrt 5$
The function is: $f(x)=(x+5i) (x-5i) (x+3)(x-1)$
Work Step by Step
We are given the function as: $f(x)=x^4-4x^3+9x^2-20x +20$
The factor theorem states that when $f(a)=0$, then we have $(x-a)$ as a factor of $f(x)$ and when $(x-a)$ is a factor of $f(x)$, then $f(a)=0$.
We factor as follows:
$x^2(x^2-4x+4)+5 (x^2-4x+4) =0 \\(x^2+5)(x^2-4x+4) =0 \\ (x^2+5)(x-2)^2=0$
By the zero product property, we have:
$x^2+5 =0 \implies x=\pm \sqrt {5} =\pm i \sqrt 5$ and $(x-2)^2 =0 \implies x=2$ (with multiplicity $2$)
Thus, we have:
The complex zeros are: $2$ with multiplicity $2, i \sqrt 5$, and $-i \sqrt 5$
The function is: $f(x)=(x+5i) (x-5i) (x+3)(x-1)$
