91Ó°ÊÓ

The given function is$f\left(x\right)=x^3-3x^2-6x+8$.

Identify any change in signs from one term to the next term in the given function.

Determine the total number of sign changes.

$$\begin{gathered} +x^3âŸ\P -3x^2 \\ -6x→+8 \end{gathered}$$

Since $f\left(x\right)$has 2 sign changes, there are two positive real zeros or no positive real zeros.

First, evaluate $f\left(-x\right)$.

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

Identify any change in signs from one term to the next term in the given function.

Determine the total number of sign changes.

$-3x^2âŸ\P +6x$

Since $f\left(-x\right)$has 1 sign change, there is one negative real zero.

List all the possible factors $p$of $a_0$.

$p=Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 8$

List all the possible factors $q$of $a_3$.

$q=Â\pm 1$

Identify all possible rational zeros $\frac{p}{q}$.

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

Perform long division to check if $x-1$is a factor of $f\left(x\right)$
.

Now, the function is written as $f\left(x\right)=\left(x-1\right)\left(x^2-2x-8\right)$.

localid="1649220997333" $\begin{array}{rcl}f\left(x\right)& =& \left(x-1\right)\left(x^2-4x+2x-8\right)\\ & =& \left(x-1\right)\left(x\left(x-4\right)+2\left(x-4\right)\right)\\ & =& \left(x-1\right)\left(x-4\right)\left(x+2\right)\end{array}$

The zeros of $f\left(x\right)$are -2, 1 and 4 .

role="math" localid="1649221048830" $f\left(x\right)=\left(x-1\right)\left(x-4\right)\left(x+2\right)$

All of the zeros have a multiplicity of 1.