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.
 A: 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
$$(x^2-x-1)^2 = 0$$
A: Hints:
Let $x-2=p.$
Then you will have a perfect square. It comes from a general fact:
"One more than a product of four consecutive positive integers is a perfect square."
Please take a look at the following link:
Prove that the product of four consecutive positive integers plus one is a perfect square
A: 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.
A: $x^4-2x^3-x^2+2x+1 = 0$
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 $\frac{1}{2}$:
$x = \frac{1}{2} + \frac{\sqrt{5}}{2} \lor x = \frac{1}{2} - \frac{\sqrt{5}}{2}$
A: obten la raiz negativa de la siguiente funcion con 4 cifras decimales.
$2x^4-2x^3+x^2+3x-4=0 \quad\quad$ newton raphson
