Re: Re: Q: Factor with Polynominals?

• To: mathgroup at smc.vnet.net
• Subject: [mg27039] Re: [mg26985] Re: Q: Factor with Polynominals?
• From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
• Date: Thu, 1 Feb 2001 03:00:26 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Actually, I just realized how to make PolynomialFactore do exactly what you
wanted (I think). Here is the improved code:

PolynomialFactor[f_, g_] :=
Block[{v},
Simplify[If[PolynomialReduce[f, g][[1, 1]] === 0,
PolynomialReduce[f, g][[2]],
PolynomialFactor[PolynomialReduce[f, g][[1, 1]], g]*g +
PolynomialReduce[f, g][[2]]] /. g -> v] /. v -> g]

Now:

In[3]:=
PolynomialFactor[3 x^2 + 3 y^2 + x + y^3 + y x^2, x^2 + y^2]

Out[3]=
2    2
x + (3 + y) (x  + y )

In[4]:=
PolynomialFactor[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2]

Out[4]=
2    2      2    2 4
y + y (x  + y ) + (x  + y )

Try it on your other examples.

--
Andrzej Kozlowski
Toyama International University
JAPAN

http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/

on 01.1.31 8:04 PM, Andrzej Kozlowski at andrzej at tuins.ac.jp wrote:

> 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.

```

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