Answer
The simplified zeros of the equation are $-1,\pm \sqrt{2}$
Work Step by Step
Let us consider the given function $ f\left( x \right)={{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2.$ The constant term is $-2$ and the leading coefficient is $1$.
The factor of the constant term is:
$-2=\pm 1,\ \pm 2$
And, the factor of the leading coefficient is
$1=\pm 1$
Therefore, the possible rational zeros of the function $ f\left( x \right)$ are
$\begin{align}
& \text{Possible rational zero}=\frac{\text{Factor of}\left( -2 \right)}{\text{Factor of 1}} \\
& =\frac{\pm 1,\ \pm 2}{\pm 1} \\
& =\pm 1,\ \pm 2
\end{align}$
Now, check whether these values $\pm 1,\ \pm 2$ are a solution of the provided function.
Let $ x=1$ and substitute this in the provided function:
$\begin{align}
& {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\
& {{1}^{4}}+2\times {{1}^{3}}-{{1}^{2}}-4\times 1-2=0 \\
& 1+2-1-4-2=0 \\
& -4\ne 0
\end{align}$
Then, $ x=1$ is not a zero of the given function $ f\left( x \right)$.
Similarly, now check for $ x=-1$:
$\begin{align}
& {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\
& {{\left( -1 \right)}^{4}}+2\times {{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}-4\times \left( -1 \right)-2=0 \\
& 1-2-1+4-2=0 \\
& 0=0
\end{align}$
Then, $ x=-1$ is a solution of the given function $ f\left( x \right)$.
Similarly, now check for $ x=2$:
$\begin{align}
& {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\
& {{2}^{4}}+2\times {{2}^{3}}-{{2}^{2}}-4\times 2-2=0 \\
& 16+16-4-8-2=0 \\
& 18\ne 0
\end{align}$
Thus, $ x=2$ is not a solution of the given function $ f\left( x \right)$.
Similarly, now check for $ x=-2$:
$\begin{align}
& {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\
& {{\left( -2 \right)}^{4}}+2\times {{\left( -2 \right)}^{3}}-{{\left( -2 \right)}^{2}}-4\times \left( -2 \right)-2=0 \\
& 16-16-4+8-2=0 \\
& 2\ne 0
\end{align}$
So, $ x=-2$ is not a solution of the given function $ f\left( x \right)$.
As shown above, $ x=-1$ is a solution of the given function.
So, $ x+1$ is a coefficient of the given function.
Find out the other coefficient by dividing the provided function by $ x+1$:
$\frac{{{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2}{x+1}={{x}^{3}}+{{x}^{2}}-2x-2$
Simplify the equation ${{x}^{3}}+{{x}^{2}}-2x-2=0$ as follows:
$\begin{align}
& {{x}^{3}}+{{x}^{2}}-2x-2=0 \\
& {{x}^{2}}\left( x+1 \right)-2\left( x+1 \right)=0 \\
& \left( {{x}^{2}}+2 \right)\left( x+1 \right)=0
\end{align}$
Therefore, the other solution of this function is $ x=-1.$
Since
$\begin{align}
& {{x}^{2}}+2=0 \\
& x=\pm \sqrt{2}
\end{align}$
Hence, the solution of this function is $-1,\ \pm \sqrt{2}$.