Answer
$1,4, -2$ and $f(x)=(x-1)(x-4)(x+2)$
Work Step by Step
We are given the function: $f(x)=x^3-3x^2-6x+8$
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 $f(x)=x^3-3x^2-6x+8$
Simplify the given function as follows:
$x^3-3x^2-6x+8=0 \\ (x-1)(x^2-2x-8)=0 \\(x-1)(x-4)(x+2)=0$
By the zero product property, we have:
$x=1, 4, -2$
So, the required zeros are: $1,4, -2$ and $f(x)=(x-1)(x-4)(x+2)$
