Re: Re: Re: Factoring question
- To: mathgroup at smc.vnet.net
- Subject: [mg35517] Re: [mg35485] Re: Re: Factoring question
- From: Andrzej Kozlowski <andrzej at yhc.att.ne.jp>
- Date: Wed, 17 Jul 2002 02:09:10 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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: [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/