Answer
$\{-1,2 \}$
Work Step by Step
Step 1. Given $f(x)=x^4-x^3+2x^2-4x-8$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm4,\pm8$
Step 2. Use synthetic division as shown in the figure to find two zero $x=-1,2$.
Step 3. Use the quotient to solve $x^2+4=0$ or $x^2=-4$, thus $x=\pm 2 i$
Step 4. Thus the real zeros are $\{-1,2 \}$