       Re: Re: factoring quartic over radicals

• To: mathgroup at smc.vnet.net
• Subject: [mg37072] Re: [mg37064] Re: factoring quartic over radicals
• From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
• Date: Tue, 8 Oct 2002 07:17:32 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Actually, including 1/GoldenRatio in the extension leads to an
unnecessarily complicated formula.
In this case there is no real need to so,  since by definition

In:=
Unevaluated[1/GoldenRatio==GoldenRatio-1]//FullSimplify

Out=
True

If one really insists on having the answer in the form proposed in
Steve's original posting one can simply do:

(Collect[#1, x] & ) /@ Factor[x^4 + x^3 + x^2 + x + 1,
Extension -> {GoldenRatio}] /. -1 + GoldenRatio ->
1/GoldenRatio

(-(-1 + x/GoldenRatio - x^2))*(1 + GoldenRatio*x + x^2)

On Monday, October 7, 2002, at 06:26 PM, Jens-Peer Kuska wrote:

> Hi,
>
> In[]:=Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio,
> 1/GoldenRatio}]
> Out[]=-((-3 - 2*x + Sqrt*x + GoldenRatio*x - 3*x^2)*
>    (3 + x + Sqrt*x + GoldenRatio*x + 3*x^2))/9
>
> Regards
>   Jens
>
> Steve Earth wrote:
>>
>> Greetings MathGroup,
>>
>> My name is Steve Earth, and I am a new subscriber to this list and
>> also a
>> new user of Mathematica; so please forgive this rather simple
>> question...
>>
>> I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into
>> Mathematica
>> and have it be able to tell me that it factors into
>>
>> (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1)
>>
>> What instructions do I need to execute to achieve this output?
>>
>> -Steve Earth
>> Harker School
>> http://www.harker.org/
>
>
>
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/

```

• Prev by Date: Re: Re: Accuracy and Precision
• Next by Date: RE: trouble with pattern matching & manipulating