Answer
$\{\frac{2}{3},-1\pm\sqrt 2 \}$
Work Step by Step
Step 1. Given $f(x)=3x^3+4x^2-7x+2$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{3},\pm\frac{2}{3}$
Step 2. Use synthetic division as shown in the figure to find a zero $x=\frac{2}{3}$.
Step 3. Use the quotient to solve $3x^2+6x-3=0$ or $x^2+2x-1=0$, thus $x=\frac{-2\pm\sqrt {4+4}}{2}=-1\pm\sqrt 2$
Step 4. Thus the real zeros are $\{\frac{2}{3},-1\pm\sqrt 2 \}$
