Answer
$\{5, -3\pm2\sqrt 3\}$
Work Step by Step
Step 1. Given $f(x)=3x^3-9x^2-31x+5$, we can list possible rational zeros as $\pm1,\pm5,\pm\frac{1}{3},\pm\frac{5}{3}$
Step 2. There are 2 sign changes in $f(x)$ indicating that there could be 2 or 0 positive real zeros.
Step 3. $f(-x)=-3x^3-9x^2+31x+5$, there is 1 sign change in $f(-x)$ indicating that there must be 1 negative real zero.
Step 4. Use synthetic division to find one real zero as shown in the figure.
Step 5. The resulting quotient gives $3x^2+6x-1=0$ which gives $x=\frac{-6\pm\sqrt {6^2+4(3)}}{2}=-3\pm2\sqrt 3$
Step 6. The zeros are $\{5, -3\pm2\sqrt 3\}$
