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The main question is how do we factorize $x^4-2yx^3+3y^3x-2y^4=0$ where $y$ is a parameter.
I thought we could use Vietta's formula and solve the following system:
$$\begin{cases} \begin{split} a+b+c+d&=2y \\ ab+bc+cd+da+bd+ac&=0 \\ abc+bcd+cda+abd&=-3y^3 \\ abcd&=-2y^4 \end{split} \end{cases}$$
where $a,b,c,d$ are the real and the imaginary roots.
Is there any fast way to solve the above system? If not how would we solve the problem?
We can exploit the homogeneity of the polynomial and (on the set where $y \neq 0$) denote $u := \frac{x}{y}$, where we see that the equation is equivalent to $$u^4 - 2 u^3 + 3 u - 2 = 0 .$$ By the Rational Root Theorem, the only possible rational roots (in $u$) are $\pm 1, \pm 2$, and checking shows that among these only $+1$ is. Polynomial long division of the l.h.s. by $u - 1$ gives $$(u - 1) (u^3 - u^2 - u + 2) = 0 .$$ If the cubic is factorable over $\Bbb Q$, it has a (rational) root, and by the R.R.T. and our previous computation, $+1$ is the only possibility, but checking shows that it is not a root, so the l.h.s is completely factored over $\Bbb Q$.
Multiplying through by $y^4$ and distributing gives that, where $y \neq 0$, we have the factorization $$(x - y) (x^3 y - x^2 y^2 - x y^3 + 2 y^3) = 0 .$$ But then this factorization holds everywhere by continuity. Since all cubics have a root in $\Bbb R$, one can factor the cubic here further yielding $$(x - y) (x - \alpha y) q(x, y) = 0$$ for some $\alpha \in \Bbb R$ and homogeneous quadratic polynomial $q$. The particular cubic in $u$ has only one real root, so this is the complete factorization over $\Bbb R$. Of course, one can factor the quadratic over $\Bbb C$ and hence write the l.h.s. of the original equation as a product of four homogeneous linear factors, namely as $$(x - y) (x - \alpha y) (x - \beta y) (x - \bar{\beta} y) = 0$$ for some $\beta \in \Bbb C - \Bbb R$. In particular, the solution set (over $\Bbb C$) is the union $$\{x = y\} \cup \{x = \alpha y\} \cup \{x = \beta y\} \cup \{x = \bar \beta y\}$$ of four (complex) lines through the origin in $\Bbb C^2$, and the solution set (over $\Bbb R$) is the union $$\{x = y\} \cup \{x = \alpha y\}$$ of two lines through the origin in $\Bbb R^2$.
