<s>[INST] Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically? Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically? Are there multiple methods for doing so?
 [/INST] If you read French, you can have a description of Ferrari's method, and links to a bunch of methods for algebraic equations (some exist in other languages).
In the present case, however, as $1$ is a root,we have a factorisation by $x-1$:
$$ x^4-x^3+2x^2-x-1=(x-1)(x^3+2x+1). $$
 So we have a cubic equation in standard form. Its discriminant is $\Delta=-4\cdot 2^3-27\cdot 1^2=-59<0$, hence it has only $1$ real root (we could as well check this point with the derivative), and we can use Cardano's method:
Set $x=u+v$. The equation rewrites as:
$$u^3+v^3+(3uv+2)(u+v)=1=0.$$
As we've  replaced $1$ unknown with $2$, we can impose a condition: $3uv+2=0$. The equation is now equivalent to the system:
$$\begin{cases}uv=-\frac23\\u^3+v^3=-1\end{cases}\iff\begin{cases}u^3v^3=-\frac8{27}\\u^3+v^3=-1\end{cases} $$
Thus the problem boils down to the standard problem on quadratic equations: find two numbers, given their sum and their product. They are the roots of the equation:
$$t^2+t-\frac8{27}=0.$$
The discriminant of this equation is $\;1+\dfrac{32}{27}=\dfrac{59}{27}\;$ so
$$u^3,v^3=\frac{-1\pm\sqrt{\frac{59}{27}}}2$$
and finally:
$$  x=u+v= \sqrt[3]{\frac{-1+\sqrt{\frac{59}{27}}}2} + \sqrt[3]{\frac{-1-\sqrt{\frac{59}{27}}}2} $$
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