Solving a biquadratic $x^{4}-2x^3 + x^2 - 2x +1 =0$ How do I find the roots of $$x^{4}-2x^3 + x^2 - 2x +1 =0$$
I am not able to find any roots by trial and error. 
 A: *

*Divide throughout by $x^2$.

*Then you have $x^{2}-2x + 1 - \frac{2}{x} + \frac{1}{x^2} = 0$

*You can re-write as $x^{2}+\frac{1}{x^2} - 2(x+\frac{1}{x}) + 1 =0$

*Then use $x^{2}+\frac{1}{x^2} = (x+\frac{1}{x})^{2} -2$ and then reduce it to a quadratic by putting $y=x+\frac{1}{x}$.
A: Hint $\ $ Exploit innate symmetry! Symmetric polynomials (palindromes)  $\rm\: = x^k\, f(x\!+\!x^{-1}),\:$ e.g.
$\rm\displaystyle\quad\begin{eqnarray} x^2\, (a\ (x+x^{-1})^2 + b\ (x+x^{-1})+c)\!  &=&\ \rm a\ (x^2+1)^2 + b\ (x^3+x) + c\ x^2 \\
&=&\ \rm a\ x^4 + b\ x^3 + (c+2a)\ x^2 + b\ x + a\\
&=&\ \rm \phantom{a}\ x^4-2\ x^3 + x^2 - 2x +1
\end{eqnarray}$  
Comparing coefficients we easily deduce that $\rm\ a = 1,\, b = -2,\, c = -1.\:$ Now the problem reduces to solving two quadratics: $\rm\:a\,z^2 + b\, z + c = 0\:$ and $\rm\:z = x + 1/x,\:$ i.e. $\rm\:x^2 - z\, x + 1 = 0.\:$
Alternatively, repeatedly subtract $\rm\ a\ (x^2 +1)^k$ and then recurse on what remains.   This amounts to a special case of Gauss's algorithm for rewriting symmetric polynomials in terms of elementary symmetric polynomials. It is one of the earliest known occurrences of Grobner basis rewriting, e.g. see p. 314 of Cox, Little, O'Shea:$\ $ Ideals, Varieties and Algorithms. In chapter 7 they present a generalization to compute the ring of invariants of a finite matrix group $\rm\: G \subset GL(n,k)\:$. 
