Answer
$\frac{2}{3},2,4$
Work Step by Step
We are given the polynomial function:
$$f(x)=-3x^3+20x^2-36x+16.$$
$\bf{Step\text{ }1}$
First we will list the possible rational zeros. The leading coefficient is $-3$ and the constant term is $16$. So the possible rational zeros are:
$$\pm 1,\pm2,\pm 4,\pm 8,\pm 16,\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{4}{3},\pm \dfrac{8}{3},\pm \dfrac{16}{3}.$$
$\bf{Step\text{ }2}$
Choose reasonable values from the list using the graph of the function. We find the reasonable values are:
$$x=\dfrac{1}{3}, x=\dfrac{2}{3}, x=2,x=\dfrac{1}{2}, x=4$$
$\bf{Step\text{ }3}$
Check the values using synthetic division until a zero is found:
Test $x=\dfrac{1}{3}$.
Because we got that the remainder is not $0$, $\frac{1}{3}$ is not a zero of $f$.
Test $x=\dfrac{2}{3}$.
Because we got that the remainder is $0$, $\frac{2}{3}$ is a zero of $f$.
$\bf{Step\text{ }4}$
Write the result of the synthetic division:
$$\begin{align*}
f(x)&=\left(x-\dfrac{2}{3}\right)(-3x^2+18x-24)\\
&=3\left(x-\dfrac{2}{3}\right)(-x^2+6x-8)\\
&=(3x-2)(-x^2+6x-8).
\end{align*}$$
$\bf{Step\text{ }5}$
Find the remaining zeros of $f$ by solving $-x^2+6x-8=0$ by factoring:
$$\begin{align*}
-x^2+6x-8&=0\\
-x^2+2x+4x-8&=0\\
-x(x-2)+4(x-2)&=0\\
(4-x)(x-2)&=0\\
4-x=0&\text{ or }x-2=0\\
x_1&=4\\
x_2&=2.
\end{align*}$$
The real zeros of $f$ are $\frac{2}{3},2,4$.
