Answer
$\{-1, -\frac{1}{2}, \frac{1}{4}, 3\}$
Work Step by Step
Step 1. List possible rational zeros as $\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{8},\pm\frac{3}{8}$
Step 2. Use synthetic division to find two zeros at $x=-1,3$ as shown in the figure.
Step 3. Use the quotient to find the other zeros: $8x^2+2x-1=0$ or $(2x+1)(4x-1)=0$, thus $x=-\frac{1}{2}, \frac{1}{4}$
Step 4. The solution set is $\{-1, -\frac{1}{2}, \frac{1}{4}, 3\}$
