## Precalculus (6th Edition) Blitzer

$x=-\frac{1}{3}, \frac{1}{2}, 1$ See graph.
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.