#### Answer

$1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$

#### Work Step by Step

We are given the polynomial function:
$$f(x)=4x^3-20x+16.$$
$\bf{Step\text{ }1}$
First we will list the possible rational zeros. The leading coefficient is $4$ and the constant term is $16$. So the possible rational zeros are:
$$\pm 1,\pm2,\pm 4,\pm 8,\pm 16,\pm \dfrac{1}{2},\pm \dfrac{1}{4}.$$
$\bf{Step\text{ }2}$
Choose reasonable values from the list using the graph of the function. We find the only reasonable value is:
$$x=1.$$
$\bf{Step\text{ }3}$
Check this value using synthetic division:
Test $x=1$.
Because we got that the remainder is $0$, $1$ is a zero of $f$.
$\bf{Step\text{ }4}$
Factor out the binomial using the result of the synthetic division:
$$\begin{align*}
f(x)&=(x-1)(4x^2+4x-16)\\
&=4(x-1)(x^2+x-4).\\
\end{align*}$$
$\bf{Step\text{ }5}$
Find the remaining zeros of $f$ by solving $x^2+x-4=0$ using the quadratic formula:
$$\begin{align*}
x&=\dfrac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}\\
&=\dfrac{-1\pm\sqrt{17}}{2}.
\end{align*}$$
The real zeros of $f$ are $1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$.