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.

Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics:
$$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \overline a)x^3 + (a\overline a + b + \overline b)x^2 + (a\overline b + \overline ab)x + b\overline b.$$
The $x^3$ coefficient is $1 = a + \overline a$, so we must have
$$a = \frac12 + si$$
for some $s ∈ ℝ$. The constant term is $1 = b\overline b$, so $b$ lies on the complex unit circle, and we must have
$$b = \cos θ + i \sin θ = \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2}i$$
where $t = \tan \frac θ2 ∈ ℝ$. The $x$ coefficient is now
$$1 = a\overline b + \overline ab = \frac{1 - 4st - t^2}{1 + t^2},$$
so we must have
$$s = -\frac t2.$$
Finally, the $x^2$ coefficient is
$$0 = a\overline a + b + \overline b = \frac{t^4 - 6t^2 + 9}{4t^2 + 4} = \frac{(t^2 - 3)^2}{4t^2 + 4},$$
so $t = \sqrt3$ (or $-\sqrt3$, which would give the same solution with the quadratics swapped).  Then $s = -\frac{\sqrt3}2$, $a = \frac{1 - i\sqrt3}2$, $b = \frac{-1 + i\sqrt3}2$, and the above factorization becomes
$$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$
