Find the complex zeros of each polynomial function f(x). Write f in factored form.

$f\left(x\right)=x^3-3x^2-6x+8$

The given function is$f\left(x\right)=(x+2)(x-1)(x-4)$

The given function has a degree 3. So, there will be a maximum of 3 complex zeroes.

We can see from the function that it has two sign variables. So, there will be two or no positive real zero.

$$\begin{gathered} f(-x)={(-x)}^3-3{(-x)}^2-6(-x)+8 \\ =-x^3-3x^2+6x+8 \end{gathered}$$

Since there is one sign change in$f(-x)$, there is one negative real zero.

We will list the integers which are factors of the constant term 8. $p=\pm 1,\pm 2,\pm 4,\pm 8$

Now, we will list the integers which are factors of the leading coefficient 1.

$q=\pm 1$

Next, list the possible potential rational zeroes which is the ratio $\frac{p}{q}$

$\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 8$

Since, there is no negative real zero, we will consider only the positive values.

Thus, the potential rational zero of function are 1,2, 4 and 8.

Using the synthetic division check whether 1 is a zero of function or not.

Thus, the depressed equation of function is

$x^2-2x-8=0$

Now, factorize the above equation by grouping.

$$\begin{gathered} (x^2+2x)+(-4x-8)=0 \\ x(x+2)+(-4)(x+2)=0 \\ (x+2)(x-4)=0 \\ x+2=0\Rightarrow x=-2 \\ x-4=0\Rightarrow x=4 \end{gathered}$$

Thus, the three complex zeroes of function are$\left\{-2,1,4\right\}$