1) I have been working with these kind of polynomials
x^2n + x^n + 1
and I realised that they are a multiple of x^2 + x + 1 but for the cases when n is a multiple of 3.
Adding and substracting x^n, I get:
x^2n + 2 x^n + 1 - x^n = (x^n +1)^2 - x^n
And, if n is an even number, n/2 is a natural number, so, I can factore the expression this way:
[x^n + 1 - x^(n/2)][x^n + 1 + x^(n/2)]
But I cant do this if n is an odd number, since I wouldnt get polynomials factors. But, dividing the original polynomial by x^2 + x + 1, I have noticed that the residual is 0 if n is not a multiple of 3. Any suggestions?
2) In these cases:
x^2n - x^n + 1, adding and substracting 3x^n, I get:
x^2n + 2 x^n + 1 - 3x^n = (x^n +1)^2 - 3x^n
And, if n is an even number, n/2 is a natural number, so, I can factore the expression this way:
[x^n + 1 - V3 x^(n/2)][x^n + 1 + V3 x^(n/2)]
Id like suggestions too for n an odd number
Thanks.
2007-08-11
12:22:30
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4 answers
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asked by
Ilusion
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in
Mathematics