<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] $x^4-2x^3-x^2+2x+1 = 0$
You should split the middle terms and try to look for a common factor.
$x^4-x^3-x^2-x^3+x^2+x-x^2+x+1 = 0$
$x^2(x^2-x-1)-x(x^2-x-1)-1(x^2-x-1) = 0$
$(x^2-x-1)^2 = 0$
Take the square root:
$x^2-x-1 = ±0$
$x^2-x-1 = 0 \lor x^2-x-1 = -0$
Add $\frac{5}{4}$:
$x^2-x+\frac{1}{4} = \frac{5}{4} \lor x^2-x+\frac{1}{4} = \frac{5}{4}$
We have two identical quadratic equations so all roots will coincide and there will be two repeated roots instead of four distinct roots.
$(x-\frac{1}{2})^2 = \frac{5}{4}$
Take the square root:
$x-\frac{1}{2} = ±\sqrt{\frac{5}{4}}$
$x-\frac{1}{2} = \sqrt{\frac{5}{4}} \lor x-\frac{1}{2} = -\sqrt{\frac{5}{4}}$
Add $\frac{1}{2}$:
$x = \frac{1}{2} + \frac{\sqrt{5}}{2} \lor x = \frac{1}{2} - \frac{\sqrt{5}}{2}$
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