factoring argument

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

User avatar
musicmunky
Posts: 34
Joined: Sun Feb 24, 2008 4:02 pm UTC
Location: West Hartford, CT
Contact:

factoring argument

Postby musicmunky » Sun Apr 20, 2008 4:37 pm UTC

So I'm trying to convince a friend of mine that I can factor 3k-1, but he doesn't believe me and I'm all hopped up on allergy meds so my factoring skillz aren't what they should be at the moment. What's the best way to factor this??
Thanks!
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
- Janos Von Neumann

Ended
Posts: 1459
Joined: Fri Apr 20, 2007 3:27 pm UTC
Location: The Tower of Flints. (Also known as: England.)

Re: factoring argument

Postby Ended » Sun Apr 20, 2008 4:50 pm UTC

3k is always odd, so 3k-1 is always even... Try expanding (1+2)k using the binomial theorem.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola

User avatar
NathanielJ
Posts: 882
Joined: Sun Jan 13, 2008 9:04 pm UTC

Re: factoring argument

Postby NathanielJ » Sun Apr 20, 2008 5:02 pm UTC

If you're thinking of something like the difference of squares technique, then no such thing exists. As Ended pointed out, a number of that form is always even, so you can always factor out a 2 (and get the remaining term from the binomial expansion), but that's as far as you can go with it in general, since in many cases (such as the k = 3 and the k = 7 cases) these are the only two factors.
Homepage: http://www.njohnston.ca
Conway's Game of Life: http://www.conwaylife.com

User avatar
SimonM
Posts: 280
Joined: Sat Jul 21, 2007 4:49 pm UTC
Location: Guernsey, CI
Contact:

Re: factoring argument

Postby SimonM » Sun Apr 20, 2008 7:21 pm UTC

NathanielJ wrote:If you're thinking of something like the difference of squares technique


Isn't this exactly the difference of two powers technique.

xa-ya = (x-y)(xa-1y0 + xa-2y1 +...)

Interesting aside.

Look to find the sum of two odd powers factorisation
mosc wrote:How did you LEARN, exactly, to suck?

User avatar
NathanielJ
Posts: 882
Joined: Sun Jan 13, 2008 9:04 pm UTC

Re: factoring argument

Postby NathanielJ » Sun Apr 20, 2008 8:13 pm UTC

SimonM wrote:
NathanielJ wrote:If you're thinking of something like the difference of squares technique


Isn't this exactly the difference of two powers technique.

xa-ya = (x-y)(xa-1y0 + xa-2y1 +...)

Interesting aside.

Look to find the sum of two odd powers factorisation


Yes of course, my point was just that there's no "non-trivial" factoring that can be done with something of the form 3k - 1, as in there's no way to factor it into something where both factors depend on k. You can always divide it by 2, but that's about it.
Homepage: http://www.njohnston.ca
Conway's Game of Life: http://www.conwaylife.com

demon
Posts: 170
Joined: Tue Feb 06, 2007 8:13 pm UTC

Re: factoring argument

Postby demon » Sun Apr 20, 2008 10:19 pm UTC

Well, you could do a tad better than that, I guess - if k is not prime, than k=m*n, so 3k-1 = 3mn-1 = (3m)n - 1n = (3m - 1)*(the sum you would expect here).
So basically, if k mod n = 0, 3n-1 divides 3k-1. It's not big result, though.

User avatar
musicmunky
Posts: 34
Joined: Sun Feb 24, 2008 4:02 pm UTC
Location: West Hartford, CT
Contact:

Re: factoring argument

Postby musicmunky » Mon Apr 21, 2008 8:08 pm UTC

Actually demon's response is perfect - I I knew you could break k up into m*n if k is not prime, but I wasn't sure of the format. The other sum is (I think) (3n+ 3n-1 + ... + 3 + 1).
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
- Janos Von Neumann

demon
Posts: 170
Joined: Tue Feb 06, 2007 8:13 pm UTC

Re: factoring argument

Postby demon » Mon Apr 21, 2008 9:28 pm UTC

Well, almost - 3mn-1n = (3m-1)(3m(n-1)+3m(n-2)+...+3m+1) I think.

Yesila
Posts: 221
Joined: Sun Dec 16, 2007 11:38 am UTC

Re: factoring argument

Postby Yesila » Mon Apr 21, 2008 9:41 pm UTC

musicmunky wrote:So I'm trying to convince a friend of mine that I can factor 3k-1, but he doesn't believe me and I'm all hopped up on allergy meds so my factoring skillz aren't what they should be at the moment. What's the best way to factor this??
Thanks!



for k=0 the problem is trivial since the factoring is 0 x anything. for k other than 0... and keeping in mind that "best" is subjective and assuming that you don't remember formulas you can always go the route of

1) define a function f(x)=x[sup]k[\sup] -1
2) notice that f(1)=0, thus (x-1) is a factor of f(x) and f(x)=(x-1)(g(x)) for some function g(x).

Take x=3 and your done with the first part of "convincing a person that it does factor. If you really want to finish the factoring now all you need to do is

3) do the long division (f(x))/(x-1). Of course this division will work much nicer (i.e. terminate) if you restrict k to be a positive integer.

Fuzz
Posts: 4
Joined: Sat Feb 09, 2008 12:35 pm UTC

Re: factoring argument

Postby Fuzz » Wed Apr 23, 2008 7:09 am UTC



Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 10 guests