Answer
$\{-3,-2\}$
Work Step by Step
Step 1. Given $f(x)=x^4+4x^3+2x^2-x+6$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$
Step 2. Use synthetic division as shown in the figure to find a zero $x=-2,-3$.
Step 3. Use the quotient and solve $x^2-x+1=0$ and there are no real zeros.
Step 4. Thus the real zeros are $\{-3,-2\}$