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