Answer
$f(x)=(x-1)(2x+1)(x^2-2)$
Real Zeros: $-\sqrt 2, \sqrt 2, 1, \dfrac{-1}{2}$ 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: $m=\pm 1, \pm 2$ and $n=\pm 1, \pm 2$
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm \dfrac{1}{2}, \pm 2$
We test with synthetic division; we will try $x -1$.
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -1 &-5&2 &2\\\hline
&2& 1 &-4 &-2\\\hline
2& 1 &-4 &-2 &0 |\ \ 0\end{array}$
Thus, we have: $f(x)=(x-1)(2x^3+x^2-4x-2)$
We test with synthetic division; we will try $x+\dfrac{1}{2}$.
$\left.\begin{array}{l}
-1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 1 &-4&-2\\\hline
&-1& 0 &2\\\hline
2& 0 &-4 &|\ \ 0\end{array}$
Thus, we have: $f(x)=(x-1)(2x+1)(x^2-2) \implies f(x)=(x-1)(2x+1)(x^2-2)$
Real Zeros: $-\sqrt 2, \sqrt 2, 1, \dfrac{-1}{2}$ and all with multiplicity $1$.
