How to find the roots of $x^6-2x^5+3x^4-2x^3+2x-1$? I want to find the roots of the sextic
$$ x^6 - 2x^5 + 3x^4 - 2x^3 + 2x - 1 $$
I have plotted the function and found two "fractional" real roots $0.5698$ and $-0.75487$. Since these are not exact roots, I can not do polynomial division to reduce the degree.
The other $4$ roots should be imaginary.  How to get those imaginary roots? Is there any roots with absolute value $> 1$?
I think we have to solve using numerical method. I appreciate your help.

Doing the same as @Martin R
$$x^6 - 2 x^5 + 3 x^4 - 2 x^3 + 2 x - 1=\left(x^3-x^2+1\right) \left(x^3-x^2+2 x-1\right)$$
$x^3-x^2+1=0$ has only one real root given by
$$x=\frac 13  \left(1-2 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{25}{2}\right)\right)\right)$$
$x^3-x^2+2 x-1=0$ has only one real root given by
$$x=\frac 13  \left( 1+2 \sqrt{5} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{11}{10 \sqrt{5}}\right)\right)\right)$$ I just followed the steps given here.
Now, you reduce the problem to two quadratic equations and obtain the analytical expressions for the complex roots.
I think that is is worth to know and practice this hyperbolic method since the probability than a cubic equation has one real root is about $0.875$.
