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RE: Re: Re: Factoring question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg35521] RE: [mg35485] Re: Re: Factoring question
*From*: "DrBob" <majort at cox-internet.com>
*Date*: Wed, 17 Jul 2002 02:09:16 -0400 (EDT)
*Reply-to*: <drbob at bigfoot.com>
*Sender*: owner-wri-mathgroup at wolfram.com
I have lots of ideas.
First of all, Mathematica 4.2 does factor the polynomial:
expr = y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2;
Factor[expr]
(-(p - y))*(q - y)*(p*q - y^2)
Secondly, you can do this manually by following steps like these:
qList = CoefficientList[expr, q]
{(-p)*y^3 + y^4, p^2*y - y^3, -p^2 + p*y}
cList = CoefficientList[(a q + b)(c q + d), q]
{b d, b c + a d, a c}
At this point, you see that a c == -p^2 + p*y, so it's natural to try
a=p and c=y-p (the other option would involve a=1) as follows:
rList = cList /. {a -> p, c -> y - p} // Expand
{b*d, (-b)*p + d*p + b*y, -p^2 + p*y}
Typically (in college algebra) we'd solve this by trial and error, but
here's a solution in Mathematica:
soln = Solve[{rList[[1]] == qList[[1]], rList[[2]] == qList[[2]]}, {b,
d}]
{{d -> p*y - y^2, b -> -y^2}, {d -> (p*y^2 - y^3)/p, b -> (-p)*y}}
(a q + b)(c q + d) /. {a -> p, c -> y - p} /. soln[[1]] // Simplify
(p - y)*(-q + y)*(p*q - y^2)
(a q + b)(c q + d) /. {a -> p, c -> y - p} /. soln[[2]] // Simplify
(p - y)*(-q + y)*(p*q - y^2)
Both solutions give the same factorization.
Here's a manual solution for b and d. The possible values for d are the
factors of:
Factor[qList[[1]]]
(-(p - y))*y^3
and those factors are
tbl=Flatten@Table[(p - y)^j y^k, {k, 0, 1}, {j, 0, 3}]
so here's the trial-and-error method:
TableForm[
Simplify[{b,
d, rList[[2]], qList[[2]]} /. {d -> #, b -> qList[[1]]/#} & /@ tbl],
TableHeadings -> {None, {b, d, rList[[2]], qList[[2]]}}]
Row 6 shows a solution, where the third and fourth columns are equal. I
could write code to pick out that row automatically (if it exists), but
you get the idea.
You could do the same thing starting with a quadratic in p instead of q.
Starting with a fourth-degree polynomial in y wouldn't work so well.
Doing all this by hand is tedious but not really difficult.
Bobby Treat
-----Original Message-----
From: Steven Hodgen [mailto:shodgen at mindspring.com]
To: mathgroup at smc.vnet.net
Subject: [mg35521] [mg35485] Re: Re: Factoring question
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: [mg35521] [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
> > >
> > >
> > >
> > >
> > >
> > >
> > >
>
>
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