Answer
$f(x)=(x+1)(2x-1)(x+\sqrt 3)(x-\sqrt 3)$
Real Zeros: $-\sqrt 3, \sqrt 3, -1, \dfrac{1}{2}$ and all with multiplicity $1$.
Work Step by Step
We see from the given polynomial function that it has at most $4$ real zeros as degree is $4$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 3$and $n=\pm 1,\pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm 3$
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm 3$
We test with synthetic division, we will try $x+1$.
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 1 &-7&-3 &3\\\hline
&-2& 1 &6 &-3\\\hline
2& -1 &-6 &3 |\ \ 0\end{array}$
Thus, we have: $f(x)=(x+1)(2x^3-x^2-6x+3)$
We test with synthetic division; we will try $x-\dfrac{1}{2}$.
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -1 &-6&3\\\hline
&1& 0 &-3\\\hline
2& 0 &-6 &|\ \ 0\end{array}$
Thus, we have: $f(x)=(x+1)(2x-1)(x^2-3) \implies f(x)=(x+1)(2x-1)(x+\sqrt 3)(x-\sqrt 3)$
Real Zeros: $-\sqrt 3, \sqrt 3, -1, \dfrac{1}{2}$ and all with multiplicity $1$.
