Answer
$x=-\frac{1}{3}, \frac{1}{2}, 1$
See graph.
Work Step by Step
Step 1. List the possible rational zeros:
$\frac{p}{q}=\pm1, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}$
Step 2. Test with synthetic division to find a zero at $x=1$, as shown in the figure.
Step 3. The function becomes:
$f(x)=(x-1)(-6x^2+x+1)=-(x-1)(3x+1)(2x-1)$
Thus the zeros are $x=-\frac{1}{3}, \frac{1}{2}, 1$
Step 4. Graph the function, as shown in the figure.
