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I have difficulties in factorizing the following polynomial in two variables : $6x^3 - 13x^2y + 4y^3 $. Thanks to a calculator I know that $6x^3 - 13x^2y + 4y^3 = (2y-3x)(2y-x)(y+2x)$, but I have no idea how to find this factorization. Is there any method to factor such polynomials ?

Thanks for your help.

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    You can use the rational root theorem, regarding this as a cubic in $x$, to find possible roots $x$ as functions of $y$. – rogerl Dec 15 '17 at 19:57
  • it might be simpler to factor $6x^3 - 13 x^2 + 4$ and then see that you can multiply the constants by $y.$ – Doug M Dec 15 '17 at 20:03

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This has homogeneity, a very strong property. Let me turn around my comment, let $$ w = y/x. $$ Divide the original by $x^3.$ You now need merely factor $$ 4 w^3 - 13 w + 6. $$ This has rational roots, and so on. Not difficult.

Will Jagy
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Let $x =ty$

\begin{eqnarray} 6x^3 - 13x^2y + 4y^3 &=& y^3(6t^3 -13t^2+4)\\ &=& y^3(6t^3 -12t^2-t^2+4)\\ &=& y^3(6t^2(t -2)-(t-2)(t+2))\\ &=& y^3(t-2)(6t^2-t-2)...\\ \end{eqnarray}

nonuser
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