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Re: Q: Factor with Polynominals?



thanks again for your answers!
unfortunately your method works with the (rather simple) examples, but it fails
with longer, more complex poynomials.
robert

Allan Hayes wrote:

> Here is a technique that works, with little thought, for both of the
> examples given so far in this thread.
>
> poly = (x^2 + y^2)^4 + y (x^2 + y^2) + y;
> poly2 = 3 x^2 + 3 y^2 + x + y^3 + y x^2;
>
> tf[u___ + a_ b_ + v___ + a_ c_ + w___] := u + a(b + c) + v + w;
> tf[z_] := z;
>
> fs = FullSimplify[poly, TransformationFunctions -> {Automatic, tf},
> ComplexityFunction -> (Length[#] &)
> ]



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