Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification 
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.


I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.
I realized there is no any rational root, by the rational root theorem.
The harder part is, WolframAlpha says the factorisation over $\Bbb Q$ is impossible.
Another solution method can be considered as the quasi-symmetric equations approach. (divide by $x^2$).
$$x^2+\frac 1{x^2}+x-\frac 2x=0$$
But the substitution $z=x+\frac 1x$ doesn't make any sense.
I want to ask the question here to find possible smarter ways to solve the quartic.

We can look for a difference of squares factorization. Completing the square gives
$$\left( x^2 + \frac{1}{2} x + c \right)^2 - \left( 2c + \frac{1}{4} \right) x^2 - (c + 2) x - (c^2 - 1)$$
and we want to find a value of $c$ such that the discriminant of the quadratic on the right is equal to zero. This gives
$$\Delta = (c + 2)^2 - 4 \left( 2c + \frac{1}{4} \right) \left( c^2 - 1) \right) = - 8c^3 + 12c + 5$$
which happily has a rational root $c = - \frac{1}{2}$ (I guess we must be essentially using the resolvent cubic here). This gives us a factorization
$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} \right)^2 + \frac{3}{4} (x - 1)^2$$
which gives a difference of squares factorization
$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} + \frac{i \sqrt{3}}{2} (x - 1) \right) \left( x^2 + \frac{1}{2} x - \frac{1}{2} - \frac{i \sqrt{3}}{2} (x - 1) \right)$$
and we can use the quadratic formula from here; if you want to know what the roots end up looking like you can ask WolframAlpha.
