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.
 A: 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
$$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$
A: You can easily observe that, the expression $(x-1)^2$ is almost included in the polynomial $P(x):=x^4+x^3-2x+1$.
Let's rewrite your polynomial as follows:
$$
\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}
$$
Based on this observation, we will represent the polynomial $P(x)$ as a "quadratic" polynomial:
$$
\begin{align}P(x):&=x^4+x^3-x^2+(x-1)^2\\
&=x^2(x^2+x-1)+(x-1)^2\\
&=x^4+x^2(x-1)+(x-1)^2\\
&=\color{red}{(x-1)^2}+\color{blue}{x^2}\color{red}{(x-1)}+\color{blue}{x^4}\end{align}
$$
Letting $x-1=u$ and writing $P(x)=0$ leads to:
$$
\begin{align}&\color{red}{u^2}+\color{blue}{x^2}\color{red}{u}+\color{blue}{x^4}=0\\
&\Delta_{\color{red}{u}}=x^4-4x^4=-3x^4\\
\implies &u_{1,2}=\frac{-x^2\pm i\sqrt 3x^2}{2}\\
\implies &u_{1,2}=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x-1=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)-x+1=0.\end{align}
$$
Now, this is obvious that the given polynomial has no real roots.
Finally, you can complete the solution by using the quadratic formula to find all complex roots.
A: Substitute $x=y+1$, then $z=y+1/y$ to get
$$y^4+5y^3+9y^2+5y+1=0$$
$$y^2((y+1/y)^2+5(y+1/y)+7)=0$$
$$z^2+5z+7=0$$
Then $z=\frac{-5\pm\sqrt{-3}}2$ and $x=\frac{z\pm\sqrt{z^2-4}}2+1$, where the four solutions are obtained from the different sign choices for $z$ and $x$.
More concretely, let $t,u=\frac{-1\pm\sqrt{-3}}2$ be the two roots of $x^2+x+1$, then
$$x^4+x^3-2x+1=(x^2-tx+t)(x^2-ux+u)$$
A: 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.
A: HINT.
$$4(x^4+x^3-2x+1)=(2x^2+x-1)^2+3(x-1)^2$$
$$=\big(2x^2+(1+i\sqrt3\,)(x-1)\big)\big(2x^2+(1-i\sqrt3)(x-1)\big),$$
DETAILS.

*

*The first quadratic term can be chosen in the form of $(2x^2+x+a)^2,$ which provides the coefficients near $x^4$ and $x^3.$


*The difference between the left part and the first quadratic term should be a full square.
A: $(x^2+e^{\theta i}x+e^{\phi i})(x^2+e^{-\theta i}x+e^{-\phi i})=x^4+2\cos\theta x^3+(1+2\cos\phi)x^2+2\cos(\theta-\phi)x+1$.
Then $\cos\theta=\frac{1}{2}$, $\cos\phi=-\frac{1}{2}$, $\cos(\theta-\phi)=-1.$
Then $\theta$ is $\pm\frac{\pi}{3}$ and $\phi$ is $\pm\frac{2\pi}{3}$.
Their difference is $\pm\pi$. Let $\theta=\frac{\pi}{3}$ and $\phi=-\frac{2\pi}{3}$
$(x^2+e^{\frac{\pi}{3}i}x+e^{-\frac{2\pi}{3} i})(x^2+e^{-\frac{\pi}{3} i}x+e^{\frac{2\pi}{3} i})=0.$
Then we solve by quadratic formula.
