91Ó°ÊÓ

The given function $f\left(x\right)=x^3+8x^2+11x-20$is of degree three, so it has at most three real zeros.

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

The factors of the constant term $-20$are

$p:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 5,Â\pm 10,Â\pm 20$

The factors of the leading coefficient $1$are

$q:Â\pm 1$

So the possible rational zeros are

$\frac{p}{q}:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 5,Â\pm 10,Â\pm 20$

The graph of the function is given as

The graph has two turning points, so it will have three real zeros.

From the graph, it appears that $1$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $1$is a zero of the function.

And the quotient $x^2+9x+20$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=x^3+8x^2+11x-20 \\ f\left(x\right)=(x+1)(x^2+9x+20) \end{gathered}$$

Equating the depressing polynomial with zero and factoring by grouping we get

$\begin{array}{rcl}{x}^2+9x+20& =& 0\\ x^2+5x+4x+20& =& 0\\ x(x+5)+4(x+5)& =& 0\\ (x+5)(x+4)& =& 0\end{array}$

So the other real zeros are

$\begin{array}{rcl}x+5& =& 0\\ x& =& -5\end{array}$

and

$\begin{array}{rcl}x+4& =& 0\\ x& =& -4\end{array}$

All the three real zeros of the function $f\left(x\right)=x^3+8x^2+11x-20$are $-5,-4,1$.

And it can be written in factored form as

$f\left(x\right)=(x+5)(x+4)(x-1)$.