<s>[INST] 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.
 [/INST] 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
$$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$
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