Answer
$-3, -\frac{5}{3},\frac{1}{2}$
Work Step by Step
We are given the polynomial function:
$$f(x)=6x^3+25x^2+16x-15.$$
$\bf{Step\text{ }1}$
First we will list the possible rational zeros. The leading coefficient is $6$ and the constant term is $-15$. So the possible rational zeros are:
$$\pm 1,\pm3,\pm 5,\pm 15,\pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{5}{2},\pm \dfrac{15}{2},\pm \dfrac{1}{3},\pm \dfrac{5}{3},\pm \dfrac{1}{6},\pm \dfrac{5}{6}.$$
$\bf{Step\text{ }2}$
Choose reasonable values from the list using the graph of the function. We find the reasonable values are:
$$x=-3, x=-\dfrac{3}{2}, x=-\dfrac{5}{3},x=\dfrac{1}{2}, x=\dfrac{1}{3}$$
$\bf{Step\text{ }3}$
Check the values using synthetic division until a zero is found:
Test $x=-3$:
Because we got that the remainder is $0$, $-3$ is a zero of $f$.
$\bf{Step\text{ }4}$
Write the result of the synthetic division:
$$\begin{align*}
f(x)&=(x+3)(6x^2+7x-5).\\
\end{align*}$$
$\bf{Step\text{ }5}$
Find the remaining zeros of $f$ by solving $6x^2+7x-5=0$ using the quadratic formula:
$$\begin{align*}
x&=\dfrac{-7\pm\sqrt{7^2-4(6)(-5)}}{2(6)}\\
&=\dfrac{-7\pm 13}{12}\\
x_1&=\dfrac{-7-13}{12}=-\dfrac{5}{3}\\
x_2&=\dfrac{-7+13}{12}=\dfrac{1}{2}.
\end{align*}$$
The real zeros of $f$ are $-3, -\frac{5}{3},\frac{1}{2}$.
