       Re: silly questions?

• To: mathgroup at smc.vnet.net
• Subject: [mg59114] Re: silly questions?
• From: Peter Pein <petsie at dordos.net>
• Date: Fri, 29 Jul 2005 00:41:51 -0400 (EDT)
• References: <dc77me\$k74\$1@smc.vnet.net> <dc9van\$cka\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```snoofly schrieb:
> Not silly at all.
> I was going to say this could be a case of one mans simple is another mans
> complex.
> One may think (x^5-32)/(x-2) is simpler than 16 + 8*x + 4*x^2 + 2*x^3 + x^4
> due to the lower number of components but experimenting below I can see that
> my theory is not the case.
>
> Messing around, I was suprised to find:
> Factor[x^4 - 16]
>
> (-2 + x)*(2 + x)*(4 + x^2)
>
> FullSimplify[(x^4 - 16)/(2 + x)]
>
> (-16 + x^4)/(2 + x)
>
> FullSimplify[(x^4 - 16)/(-2 + x)]
>
> (2 + x)*(4 + x^2)
>
> Why does the first FullSimplify not return (-2 + x)*(4 + x^2) in similar
> style to the second? Clearly this must be simpler than the first (which
> involves same # components but at higher orders). Additionally, if the
> second can be done, why is that the algorithm misses on the first almost
> identical case?
>
...

>
> "Kent Holing" <KHO at statoil.com> wrote in message
> news:dc77me\$k74\$1 at smc.vnet.net...
>
>>Why does not (x^5-32)/(x-2)//FullSimplify in Mathematica  work?
>>Compare with Factor[x^5-32]//InputForm which returns (-2 + x)*(16 + 8*x +
>>4*x^2 + 2*x^3 + x^4).
>>So why does not the first command just return 16 + 8*x + 4*x^2 + 2*x^3 +
>>x^4?
>>As in a factorization above, how is the easiest way to pick automatically
>>(by a function) the factors of say degree >=2,  if any ?
>>
>>Kent Holing
>>
>
>
Hi,

if you simply try Factor[(x^4-16)/(x-#)]&/@{-2,2}, Mathematica returns
the canceled polynomials, you want. I _guess_ Factor is not among the
TransformationFunctions (see help), because it /can/ become very time
consuming.

--
Peter Pein
Berlin

```

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