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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?

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    $\begingroup$ Is $y$ a constant? Do we have to solve for $x$? $\endgroup$ – SchrodingersCat Jan 15 '16 at 16:17
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    $\begingroup$ for $y\ne 0$ we can write $$\left(\frac{x}{y}\right)^4-2\left(\frac{x}{y}\right)^3+3\frac{x}{y}-2=0$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 15 '16 at 16:18
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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 $\Bbb R^2$.

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    $\begingroup$ In case OP is interested in the other roots then they can be found using Cardano's, but it's very messy and the result is not very pretty: $\frac{1}{3} \left(1-\frac{4}{\alpha}-\alpha\right)$ and $\frac{6 + \frac{4}{\alpha}+3\alpha}{18} \pm i \frac{\sqrt{3}}{6} \left[\frac{4}{\alpha} - \alpha\right]$ where $\alpha = \sqrt[3]{\frac{43}{2}-\frac{3\sqrt{177}}{2}}$. $\endgroup$ – Kibble Jan 15 '16 at 17:04

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