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