## factoring argument

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musicmunky
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### factoring argument

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!
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Ended
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### Re: factoring argument

3k is always odd, so 3k-1 is always even... Try expanding (1+2)k using the binomial theorem.
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NathanielJ
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### Re: factoring argument

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.
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SimonM
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### Re: factoring argument

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
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NathanielJ
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### Re: factoring argument

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.
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demon
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### Re: factoring argument

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.

musicmunky
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### Re: factoring argument

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."
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demon
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### Re: factoring argument

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

Yesila
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### Re: factoring argument

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
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