<s>[INST] Solve the equation $x^4+(x-1)(x^2-2x+2)=0$ With $x \in \mathbb{R}$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$

Idea $1$ : $(x^2-x+1)(x^2+2x-2)=0$
Idea $2$ :$(2x^2+x-1)^2=(3x-3)^2$
Idea $3$ :$<=>x^4(x-1)x^2-2(x-1)^2=0$
Let $y=x-1 \dots$
I need another way because I am collecting methods of solving with quartic equations
 [/INST] 
Solve $x^4+(x-1)(x^2-2x+2)=0$

You can solve this using Newton-Raphson :
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
Where \begin{align}&f(x)=x^4+x^3-3x^2+4x-2\\
&f'(x)=4x^3+3x^2-6x+4\\\end{align}
To guess $x_0$ :
\begin{align}&f(1)>0
\\&f(0)<0
\\\end{align}
So $x_0=1$ :
\begin{align}&x_1=0.5\\
&x_2=0.8\\
&x_3\approx0.6765\\
& \dots \\ 
&\dots \\ 
&\dots \\
&x_{22}\approx0.73205  \\\end{align}
The first Solution is $\boxed{x\approx0.73205}$ now Apply long division  :
$$\frac{x^4+x^3-3x^2+4x-2}{x-0.73205 }=x^3+1.73205 x^2-1.73205 x+2.73205$$
Now solve this equation with the same method you should get the $2$nd Solution if there is one
$$x^3+1.73205x^2-1.73205 x+2.73205 \approx 0$$
If there is a Solution apply long division again to find another one until you come to a function that doesn't have Solutions where $x \in\mathbb{R}$.
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