Answer
$f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$
Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$
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 $ 2,\qquad \pm 1,\pm 2,$
and $q$ must be a factor of $ 2,\qquad \pm 1,\pm 2$
$\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm 2$
Testing with synthetic division, try $x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &-1 &-5 &2 &2 \\\hline
&2 &1 &-4 & -2 \\\hline
2& 1 &-4 &-2& |\ \ 0 \end{array}$
$f(x)=(x-1)(2x^{3}+x^{2}-4x-2)$
Try synthetic division again, $ x+\displaystyle \frac{1}{2}$
$\left.\begin{array}{l}
-1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &1 &-4 &-2 \\\hline
&-1 &0 & +2 \\\hline
2 & 0 &-4 & |\ \ 0 \end{array}$
$f(x)=(x-1)(x+\displaystyle \frac{1}{2})(2x^{2}-4)=(x-1)(x+\frac{1}{2})\cdot 2(x^{2}-2)$
$=(x-1)(2x+1)(x^{2}-2)$
Solving $x^{2}-2=0,$
$x=\pm\sqrt{2}$, we have the other two zeros.
$f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$
Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$
all with multiplicity 1.