Let $$f=x^9-x^6+4x^5y+2x^3y^2-y^4$$ I would like to factorize $f$ into form: $$(y-F_1(x))\cdots(y-F_k(x))$$ over complex numbers. How can I do it with Mathematica?
1 Answers
Since we can consider $f$ as a $4-$th order polynomial (with respect to $y$) we can always factorize it as you claim in terms of radicals, but for the sake of simplicity let's start writing it symbolically in terms of the Root objects defining $f$ to be p[x,y]:
p[x_, y_] := x^9 - x^6 + 4 x^5 y + 2 x^3 y^2 - y^4
pf[x_, y_] = -Times @@ (y - (y /. {ToRules @ Reduce[p[x, y] == 0, {x, y}]}))
-((y - Root[x^6 - x^9 - 4 x^5 #1 - 2 x^3 #1^2 + #1^4 &, 1]) (y - Root[x^6 - x^9 - 4 x^5 #1 - 2 x^3 #1^2 + #1^4 &, 2]) (y - Root[x^6 - x^9 - 4 x^5 #1 - 2 x^3 #1^2 + #1^4 &, 3]) (y - Root[x^6 - x^9 - 4 x^5 #1 - 2 x^3 #1^2 + #1^4 &, 4]))
We can simplify this form to its original form:
% // Simplify
% == p[x, y]
-x^6 + x^9 + 4 x^5 y + 2 x^3 y^2 - y^4True
We can use the above factorized form pf[x,y] consequently in computations. It might be advantageous to observe how the roots depend on the x parameter with the following animation:
anim =
Table[
GraphicsRow[
Table[
ContourPlot[ g[ p[x, u + I w]], {u, -10, 10}, {w, -10, 10},
Epilog -> {Blue, PointSize[0.03], Point[ Table[
{Re @ #, Im @ #}& @ Root[x^6 - x^9 - 4 x^5 #1 - 2 x^3 #1^2 + #1^4 &, m],
{m, 4}]]},
ColorFunction -> "StarryNightColors", ImageSize -> 400],
{g, {Re, Im}}]],
{x, -2.5, 2.5, 0.1}];
ListAnimate[ anim, Paneled -> False]

Alternatively we could exploit the Quartics -> False option of Solve (mind we have used only y as a variable, otherwise we would get a warning):
-Times @@ (y - (y /. Solve[ p[x, y] == 0, y, Quartics -> False]))
It's traditional form in terms of radicals is slightly involved:
ToRadicals[ -pf[x, y]] // TraditionalForm

This is the reason why we need sometimes options like Quartics or Cubics. For more information on roots see e.g. How do I work with Root objects?, how you could exploit Factor: Factoring polynomials to factors involving complex coefficients.