Answer
$\{-2,-1,2,2 \}$, $f(x)=(x+2)(x+1)(x-2)^2$
Work Step by Step
Step 1. Given $f(x)=x^4-x^3-6x^2+4x+8$, 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 $x=2,-1$.
Step 3. Use the quotient to solve $x^2-4=0$ or $x^2=4$, thus $x=-2,2$
Step 4. Thus the real zeros are $\{-2,-1,2,2 \}$ and we can factor the function as $f(x)=(x+2)(x+1)(x-2)^2$
