Answer
$f(x)=(x+1)(2x-1)(x+\sqrt{3})(x-\sqrt{3})$
Zeros: $-\displaystyle \sqrt{3},\ \ -1,\ \ \frac{1}{2},\ \ \sqrt{3},$
all with multiplicity 1.
Work Step by Step
$f(x)$ has at most $4$ real zeros, as the degree is $4$.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $ 3,\qquad \pm 1,\pm 3,$
and $q$ must be a factor of $ 2,\qquad \pm 1,\pm 2$
$\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm\frac{3}{2},\pm 3$
Testing with synthetic division, try $x+1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &1 &-7 &-3 &3 \\\hline
&-2 &1 &6 & -3 \\\hline
2&-1 &-6 &3& |\ \ 0 \end{array}$
$f(x)=(x+1)(2x^{3}-x^{2}-6x+3)$
Try synthetic division again, $ x-\displaystyle \frac{1}{2}$
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &-1 &-6 &3 \\\hline
&1 &0 & -3 \\\hline
2 & 0 &-6 & |\ \ 0 \end{array}$
$f(x)=(x+1)(x-\displaystyle \frac{1}{2})(2x^{2}-6)=(x+1)(x-\frac{1}{2})\cdot 2(x^{2}-3)$
$=(x+1)(2x-1)(x^{2}-3)$
Solving $x^{2}-3=0,$
$x=\pm\sqrt{3}$, we have the other two zeros.
$f(x)=(x+1)(2x-1)(x+\sqrt{3})(x-\sqrt{3})$
Zeros: $-\displaystyle \sqrt{3},\ \ -1,\ \ \frac{1}{2},\ \ \sqrt{3},$
all with multiplicity 1.