<s>[INST] How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0  $$ $$x(x-2)(x^2-1)+1=0   $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
 [/INST] Another way to notice the factorization
$$x^4-2x^3-x^2+2x+1=0$$
Since $x=0$ is not the root of the equation, divide by $x^2$ to get
$$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$
Rewrite it as
$$x^2+\frac{1}{x^2} - 2\left(x-\frac{1}{x}\right) - 1 = 0$$
or
$$\left(x-\frac1x\right)^2 + 2 - 2\left(x-\frac{1}{x}\right) - 1 = 0$$
Substitute $t = x - 1/x$
$$t^2 + 2 - 2t - 1 = 0\\
t^2 - 2t + 1 = 0 \\ (t-1)^2 = 0$$
Substitute back to get the final result
$$\left(x - \frac{1}{x} - 1\right)^2 = 0$$
which says
$$(x^2-x-1)^2 = 0$$
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