Answer
The complex zeros are: $\dfrac{1}{2}$ with multiplicity $2$, $-2$ with multiplicity $1$.
The function is: $f(x)=4(x-\dfrac{1}{2})^2 (x+2)$
Work Step by Step
We are given the function as: $f(x)=4x^3+4x^2-7x+2$
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 can see that $f(-2)=0$. This means that $x+2$ is a factor of $f(x)=4x^3+4x^2-7x+2$ .
Simplify the given function as follows:
$4x^3+4x^2-7x+2 \\ (x+2)(4x^2-4x+1)=0 \\ (x+2)(2x-1)(2x-1)=0$
By the zero product property, we have:
$x=\dfrac{1}{2}, \dfrac{1}{2}$
Thus, we have:
The complex zeros are: $\dfrac{1}{2}$ with multiplicity $2$, $-2$ with multiplicity $1$.
The function is: $f(x)=4(x-\dfrac{1}{2})^2 (x+2)$