Answer
$x\displaystyle \in\{ -1, 2-\sqrt 2, 2+\sqrt 2, 2\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^5-4x^4-x^{3}+10x^{2}+2x-4$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 4$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}|& 1 & -4 & -1 & 10 & 2& -4\\
& & -1& 5& -4& -6& 4\\
& -- & -- & -- & --\\
& 1 & -5 & 4& 6& -4 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^4-5x^3+4x^2+6x-4)$
Try for $x=-1$:
$\begin{array}{lllll}
\underline{-1}|& 1&-5 & 4 & 6 & -4\\
& & -1& 6& -10& 4\\
& -- & -- & -- & --\\
& 1 & -6& 10& -4 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)^2(x^3-6x^2+10x-4)$
Try $x=2$:
$\begin{array}{lllll}
\underline{2}|& 1 & -6 & 10 & -4\\
& & 2& -8& 4\\
& -- & -- & -- & --\\
& 1 & -4& 2 & |\underline{0}
\end{array}$
$2$ is a zero,
$f(x)=(x+1)^2(x-2)(x^2-4x+2)$
Solving for the trinomial, using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
in this case, $x^2-4x+2$, $x=\frac{4\pm 2\sqrt 2}{2}=2\pm\sqrt 2$
Therefore, the real zeros of the equations are:
$x\displaystyle \in\{ -1, 2-\sqrt 2, 2+\sqrt 2, 2\}$
