Re: Re: Re: Re: Factoring question
- To: mathgroup at smc.vnet.net
- Subject: [mg35581] Re: [mg35517] Re: [mg35485] Re: Re: Factoring question
- From: Andrzej Kozlowski <andrzej at bekkoame.ne.jp>
- Date: Fri, 19 Jul 2002 06:09:29 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Of course if If you really insist on solving this problem by grouping
and factoring it is also quite simple:
Consider the expression:
Out[14]=
(-p^2)*q^2 + p^2*q*y + p*q^2*y - p*y^3 - q*y^3 + y^4
group it as follows:
(y^4 - p*y^3) + (p*q^2*y - p^2*q^2) + (p^2*q*y - q*y^3)
factor out y-p:
((y - p))*y^3 + p*q^2*(y - p) - (y - p)*y*(p + y)*q
hence
(y-p)(y^3 + p*q^2 -y*(p + y)*q)
re-group as
(y-p)((y^3-y^2 q) + (p q^2-y pq))
factor out y-q:
(y-p)((y-q)(y^2 -(y-p)pq) hence
(y-p)(y-q)(y^2-pq)
Andrzej Kozlowski
On Wednesday, July 17, 2002, at 02:09 AM, Andrzej Kozlowski wrote:
> Mathematica does not try to use any term groupings or re-arrangings that
> people usually try, because for computers much more direct methods are
> much faster and also guarantee getting the answer. In this case the
> expression is a quartic, so Mathematica can solve it in radicals using
> the formula of Cardano. The easiest way for a human who can't remember
> such complicated things is to notice that p and q are obviously roots
> of y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2==0. Thus (x-p)(x-q)
> must be a factor a factor. Once you know that the of the problem is
> trivial. Using Mathematica you can finish it in lots of ways, e.g.
>
> In[1]:=
> Cancel[((-p^2)*q^2 + (p^2*q + p*q^2)*y - (p + q)*y^3 + y^4)/
> ((-p + y)*(-q + y))]
>
> Out[1]=
> (-p)*q + y^2
>
> or
>
> In[2]:=
> PolynomialQuotient[y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y -
> p^2*q^2, (y - p)*(y - q), y]
>
> Out[21]=
> (-p)*q + y^2
>
>
> or
>
> In[3]:=
> SolveAlways[y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y -
> p^2*q^2 == (a*y^2 + b*y + c)*((y - p)*(y - q)), y]
>
> Out[3]=
> {{c -> (-p)*q, a -> 1, b -> 0}, {c -> 0, a -> 1, b -> 0,
> p -> 0}, {c -> 0, a -> 1, b -> 0, q -> 0}}
>
> The last method is equivalent to what is sometimes called "comparing
> coefficients".
>
>
> On Tuesday, July 16, 2002, at 04:49 AM, Steven Hodgen wrote:
>
>> Actually, I never posted the actual problem, but here it is:
>>
>> y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2
>>
>> This factors to:
>>
>> (y^2 - p*q)*(y - p)*(y - q)
>>
>> Which I cannot get. I've tried over an over again, and I'm the best
>> factorer in my class, my instructor can't seem to get this either,
>> hence my
>> hope that Mathematica could do it, and it certainly can no problem, but
>> it
>> just doesn't show how. So, any ideas?
>>
>> --Steven
>>
>> "Ken Levasseur" <Kenneth_Levasseur at uml.edu> wrote in message
>> news:agrjfj$g9n$1 at smc.vnet.net...
>>> Steve:
>>>
>>> I assume that the problem was to solve x^7 + x^5 + x^4 + -x^3 + x +
>>> 1=0.
>> If
>>> so, one of the basic factoring theorems is that if a polynomial over
>>> the
>>> integers like this one has a rational root r/s, then r must divide
>>> the
>>> constant term and s must divide the leading coefficient. So in this
>> problem,
>>> +/-1 are the only possible rational roots and so the (x+1) factor
>>> would be
>>> found this way. I'm sure that Mathematica checks this almost
>> immediately.
>>> As for the remaining 6th degree factor, I'm not certain how
>>> Mathematica
>>> proceeds, but if you plot it, it clearly has no linear factors.
>>>
>>> Ken Levasseur
>>>
>>>
>>> Steven Hodgen wrote:
>>>
>>>> "DrBob" <majort at cox-internet.com> wrote in message
>>>> news:agbfhl$je9$1 at smc.vnet.net...
>>>>> Factor[x^7 + x^5 + x^4 + -x^3 + x + 1] // Trace
>>>>
>>>> This doesn't do it. It only traces the initial evaluation, and then
>> simply
>>>> displays the factored result with no intermediate factoring steps.
>>>>
>>>> Thanks for the suggestion though.
>>>>
>>>>>
>>>>> Bobby
>>>>>
>>>>> -----Original Message-----
>>>>> From: Steven Hodgen [mailto:shodgen at mindspring.com]
To: mathgroup at smc.vnet.net
>>>>> Subject: [mg35581] [mg35517] [mg35485] Factoring question
>>>>>
>>>>> Hello,
>>>>>
>>>>> I just purchased Mathematica 4.1. I'm taking precalculus and wanted
>> to
>>>> try
>>>>> a tough factoring problem, since the teacher couldn't do it either.
>>>>> Mathematica get's the correct answer, but I'm interrested in seeing
>> how it
>>>>> got there. Is there a way to turn on some sort of trace feature
>>>>> where
>> it
>>>>> shows each step it used to get the the final result?
>>>>>
>>>>> Thanks!
>>>>>
>>>>> --Steven
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>
>>>
>>
>>
>>
>>
>>
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
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