Answer
Zeros: $\displaystyle \ \ \frac{1}{2}$, with multiplicity 1.
$f(x)= (2x-1)(x^{2}+1)$
Work Step by Step
$f(x)$ has at most 3 real zeros, as the degree is 3.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $-1,\qquad \pm 1,$
and $q$ must be a factor of $ 2,\qquad \pm 1,\pm 2$
$\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2}$
Testing with synthetic division, try $x-\displaystyle \frac{1}{2}$
$\left.\begin{array}{l}
\displaystyle \frac{1}{2} \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &-1 &2 &-1\\\hline
&1 &0 &1 \\\hline
2&0 &2& |\ \ 0 \end{array}$
$f(x)=(x-\displaystyle \frac{1}{2})(2x^{2}+2)=(x-\frac{1}{2})\cdot 2(x^{2}+1)$
$= (2x-1)(x^{2}+1)$
The binomial $x^{2}+1$ has no real zeros, so we can not factor any further.
Zeros: $\displaystyle \ \ \frac{1}{2}$, with multiplicity 1.
$f(x)= (2x-1)(x^{2}+1)$
