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Re: Re: Q: Factor with Polynominals?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg27038] Re: [mg26985] Re: Q: Factor with Polynominals?
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Thu, 1 Feb 2001 03:00:25 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
I have found a bit of time to consider your problem more carefully and I
concluded that while in principle what you want done is quite easy, there
seems to be an ambiguity in what you call "factoring a polynomial".
Here is the function which will produce the output you wanted in the first
case, while in your second case it will essentially the same output:
PolynomialFactor[f_, g_] :=
If[PolynomialReduce[f, g][[1, 1]] === 0, PolynomialReduce[f, g][[2]],
PolynomialFactor[PolynomialReduce[f, g][[1, 1]], g]*g +
PolynomialReduce[f, g][[2]]]
Here is what happens with your two examples:
In[2]:=
PolynomialFactor[3 x^2 + 3 y^2 + x + y^3 + y x^2, x^2 + y^2]
Out[2]=
x + (3 + y)*(x^2 + y^2)
This is just what you wanted.
In[3]:=
PolynomialFactor[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2]
Out[3]=
2 2 2 2 3
y + (x + y ) (y + (x + y ) )
This looks somewhat different, but actually that's because it is factored
more deeply than you asked for. This sort of problem will occur in other
examples, e.g.
In[15]:=
PolynomialFactor[(1 + x^2)^2*(x + 3) + (1 + x^2)*(x - 3), 1 + x^2]
Out[15]=
2 2
(1 + x ) (-3 + x + (3 + x) (1 + x ))
Again, this is factored further than is the case in your examples. One can
probably fix it with some further programming but quite frankly I do not
think it is worth the effort, having got to this point it should be easy to
re-arrange things by hand.
--
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/
on 01.1.31 1:22 PM, Robert at robert.schuerhuber at gmx.at wrote:
>
>
> 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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