Answer
$\{1,\frac{5\pm\sqrt {61}}{6}\}$
Work Step by Step
Step 1. Rewrite the equation as $3x^3-8x^2+2x+3=0$, list the potential rational zeros as $\pm1,\pm3,\pm\frac{1}{3}$.
Step 2. Use synthetic division to find a real zero $x=1$ as shown in the figure.
Step 3. Use the quotient and solve $3x^2-5x-3=0$ to get $x=\frac{5\pm\sqrt {25+36}}{2(3)}=\frac{5\pm\sqrt {61}}{6}$
Step 4. Thus the zeros are $\{1,\frac{5\pm\sqrt {61}}{6}\}$
