<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] General solution:
I will solve a specific quartic equation that is a specific case of a general quartic equation.

Let, $a≠0,~ b≠0$,  then we have
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x^2+\frac e{ax^2}+\frac ba x+\frac d{ax}+\frac ca=0$$
$$\begin{align}&x^2+\frac ea \times \frac 1{x^2}+\frac ba \left(x+\frac {d}{bx}\right)+\frac ca=0&\end{align}$$
This quartic equation can be directly converted to the quadratic equation in the case below, avoiding the cubic equation.
$$\begin{align}&x+\frac {d}{bx}=t \\
\implies &t^2=x^2+\frac{d^2}{b^2}\times \frac {1}{x^2}+\frac{2d}{b}\end{align}$$
Then, if $$\frac ea =\frac{d^2}{b^2}$$
We have
$$t^2-\frac{2d}{b}+\frac ba t+\frac ca=0$$
$$t^2+\frac ba t+\left(\frac ca-\frac{2d}{b}\right)=0$$
The last equation is a quadratic equation.
After solving quadratic, you wil get
$$x+\frac {d}{bx}=t$$
$$bx^2-btx+d=0$$
The last equation is also a quadratic equation.

In your case we have
$$a=1,b=-2,c=-1, d=2, e=1$$
This means
$$\frac ea =\frac{d^2}{b^2}$$
holds.
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