91Ó°ÊÓ

The given function $f\left(x\right)=2x^3-13x^2+24x-9$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 $-9$are

$p:Â\pm 1,Â\pm 3,Â\pm 9$

The factors of the leading coefficient $2$are

$q:Â\pm 1,Â\pm 2$

So the possible rational zeros are

$\frac{p}{q}:Â\pm 1,Â\pm 3,Â\pm 9,Â\pm \frac{1}{2},Â\pm \frac{3}{2},Â\pm \frac{9}{2}$

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 $3$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $3$is a zero of the function.

And the quotient $2x^2-7x+3$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=2x^3-13x^2+24x-9 \\ f\left(x\right)=(x-3)(2x^2-7x+3) \end{gathered}$$

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

$\begin{array}{rcl}2x^2-7x+3& =& 0\\ 2x^2-6x-x+3& =& 0\\ 2x(x-3)-1(x-3)& =& 0\\ (x-3)(2x-1)& =& 0\end{array}$

So the other zeros are

$\begin{array}{rcl}x-3& =& 0\\ x& =& 3\end{array}$

and

$\begin{array}{rcl}2x-1& =& 0\\ 2x& =& 1\\ x& =& \frac{1}{2}\end{array}$

All the three real zeros of the function $f\left(x\right)=2x^3-13x^2+24x-9$are $\frac{1}{2},3,3$.

And it can be written in factored form as$f\left(x\right)=(2x-1)(x-3)(x-3)$.