91Ó°ÊÓ

The given function$f\left(x\right)=x^4-x^3-6x^2+4x+8$is of degree four, so it has at most four real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $8$are

$p:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 8$

The factors of the leading coefficient $1$are

$q:Â\pm 1$

So the possible rational zeros are

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

The graph of the function is given as

So the graph has three turning points and thus has four real roots.

From the graph, it appears that $-2$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $-2$is a zero of the function.

And the quotient $x^3-3x^2+4$is the depressed polynomial. So the given function is factored as

localid="1646116620169" $f\left(x\right)=(x+2)(x^3-3x^2+4)$

From the graph, it can be seen that $2$may also be a factor. On performing synthetic division with the depressed polynomial we get

As the remainder is zero so $2$is a zero of the function.

And the quotient $x^2-x-2$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=(x+2)(x^3-3x^2+4) \\ f\left(x\right)=(x+2)(x-2)(x^2-x-2) \end{gathered}$$

Equate the depressed polynomial equal to zero and factor by grouping.

$\begin{array}{rcl}{x}^2-x-2& =& 0\\ x^2-2x+x-2& =& 0\\ x(x-2)+(x-2)& =& 0\\ (x-2)(x+1)& =& 0\end{array}$

So the other zeros are

$\begin{array}{rcl}x-2& =& 0\\ x& =& 2\end{array}$

and

role="math" $\begin{array}{rcl}x+1& =& 0\\ x& =& -1\end{array}$

And the complete factored form is$f\left(x\right)=(x+2)(x-2)(x-2)(x+1)$