So I'm trying to convince a friend of mine that I can factor 3^{k}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!
factoring argument
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 musicmunky
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factoring argument
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
 Janos Von Neumann
 Janos Von Neumann

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Re: factoring argument
3^{k} is always odd, so 3^{k}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.
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.
x^{a}y^{a} = (xy)(x^{a1}y^{0} + x^{a2}y^{1} +...)
Interesting aside.
Look to find the sum of two odd powers factorisation
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 NathanielJ
 Posts: 882
 Joined: Sun Jan 13, 2008 9:04 pm UTC
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.
x^{a}y^{a} = (xy)(x^{a1}y^{0} + x^{a2}y^{1} +...)
Interesting aside.
Look to find the sum of two odd powers factorisation
Yes of course, my point was just that there's no "nontrivial" factoring that can be done with something of the form 3^{k}  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.
Re: factoring argument
Well, you could do a tad better than that, I guess  if k is not prime, than k=m*n, so 3^{k}1 = 3^{mn}1 = (3^{m})^{n}  1^{n} = (3^{m}  1)*(the sum you would expect here).
So basically, if k mod n = 0, 3^{n}1 divides 3^{k}1. It's not big result, though.
So basically, if k mod n = 0, 3^{n}1 divides 3^{k}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) (3^{n}+ 3^{n1} + ... + 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
 Janos Von Neumann
Re: factoring argument
Well, almost  3^{mn}1^{n} = (3^{m}1)(3^{m(n1)}+3^{m(n2)}+...+3^{m}+1) I think.
Re: factoring argument
musicmunky wrote:So I'm trying to convince a friend of mine that I can factor 3^{k}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 (x1) is a factor of f(x) and f(x)=(x1)(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))/(x1). Of course this division will work much nicer (i.e. terminate) if you restrict k to be a positive integer.
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