Answer
zeros:$\qquad -1,\ 2, \ \ 2\pm\sqrt{2}$
Work Step by Step
$P(x)=x^{5}-4x^{4}-x^{3}+10x^{2}+2x-4,$
($3$ sign variations, we expect $3$ or $1$ positive real zero)
$P(-x)=-x^{5}-4x^{4}+x^{3}+10x^{2}-2x-4,$
($2$ sign variations, we expect $2$ or $0$ negative real zeros)
$a_{0}=-4 \quad $p: $\pm 1,\pm 2,\pm 4$
$a_{n}=1,\qquad $q: $\pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm 2$...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 1 & -4 & -1 & 10 & 2 & -4 & \\
& & -1 & 5 & -4 & -6 & 4 & \\
\hline & 1 & -5 & 4 & 6 & -4 & 0 & \\
& & & \swarrow & & & & \\
-1 & 1 & -5 & 4 & 6 & -4 & & \\
& & -1 & 6 & -10 & 4 & & \\
\hline & 1 & -6 & 10 & -4 & 0 & & \\
& & &\swarrow & & & & \\
2 & 1 & -6 & 10 & -4 & & & \\
& & 2 & -8 & 4 & & & \\
\hline & 1 & -4 & 2 & 0 & & & \\
\end{array}$
$P(x)=(x+1)^{2}(x-2)(x^{2}-4x+2)$
Find the zeros of the last factor with the quadratic formula
$x=\displaystyle \frac{4\pm\sqrt{(4)^{2}-4(1)(2)}}{2(1)}=\frac{4\pm\sqrt{8}}{2}=\frac{4\pm 2\sqrt{2}}{2}=2\pm\sqrt{2}$
zeros:$\qquad -1,\ 2, \ \ 2\pm\sqrt{2}$
