Answer
$-1, \frac{3}{2},\frac{5}{2}$
Work Step by Step
We are given the polynomial function:
$$f(x)=4x^3-12x^2-x+15.$$
$\bf{Step\text{ }1}$
First we will list the possible rational zeros. The leading coefficient is $4$ and the constant term is $15$. So the possible rational zeros are:
$$\pm 1,\pm3,\pm 5,\pm 15,\pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{5}{2},\pm \dfrac{15}{2},\pm \dfrac{1}{4},\pm \dfrac{3}{4},\pm \dfrac{5}{4},\pm \dfrac{15}{4}.$$
$\bf{Step\text{ }2}$
Choose reasonable values from the list using the graph of the function. We find the reasonable values are:
$$x=-1, x=-\dfrac{3}{4}, x=\dfrac{3}{2},x=\dfrac{5}{4}, x=\dfrac{5}{2}$$
$\bf{Step\text{ }3}$
Check the values using synthetic division until a zero is found:
Test $x=-1$:
Because we got that the remainder is $0$, $-1$ is a zero of $f$.
$\bf{Step\text{ }4}$
Write the result of the synthetic division:
$$\begin{align*}
f(x)&=(x+1)(4x^2-16x+15).\\
\end{align*}$$
$\bf{Step\text{ }5}$
Find the remaining zeros of $f$ by solving $4x^2-16x+15=0$ using the quadratic formula:
$$\begin{align*}
x&=\dfrac{-(-16)\pm\sqrt{(-16)^2-4(4)(15)}}{2(4)}\\
&=\dfrac{16\pm 4}{8}\\
x_1&=\dfrac{16-4}{8}=\dfrac{3}{2}\\
x_2&=\dfrac{16+4}{8}=\dfrac{5}{2}.
\end{align*}$$
The real zeros of $f$ are $-1, \frac{3}{2},\frac{5}{2}$.
