Answer
$\{-1, \frac{1}{2}, \pm \sqrt 3 \}$, $f(x)=(x+1)(2x-1)(x+\sqrt 3)(x-\sqrt 3)$
Work Step by Step
Step 1. Given $f(x)=2x^4+x^3-7x^2-3x+3$, list possible rational zeros as $\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
Step 2. Use synthetic division as shown in the figure to find two zero $x=-1,\frac{1}{2}$.
Step 3. Use the quotient to solve $2x^2-6=0$ or $x^2=3$, thus $x=\pm \sqrt 3$
Step 4. Thus the real zeros are $\{-1, \frac{1}{2}, \pm \sqrt 3 \}$ and we can factor the function as $f(x)=(x+1)(2x-1)(x+\sqrt 3)(x-\sqrt 3)$
