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Math Help - [SOLVED] Factor x^n - 1

  1. #1
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    [SOLVED] Factor x^n - 1

    Factor x^n - 1 for n = 6, 7, 8,..., 22.

    Do I need to show the answer for each n? What is the proper way to show this?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by john_n82 View Post
    Factor x^n - 1 for n = 6, 7, 8,..., 22.

    Do I need to show the answer for each n? What is the proper way to show this?
    x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1}).

    So x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5).

    I don't know if that's what you're looking for.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1}).

    So x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5).

    I don't know if that's what you're looking for.
    because the question askes me to find x^n - 1 for n = 6, 7, 8, ..., 22 so I think I have to do for each n.
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  4. #4
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    Quote Originally Posted by john_n82 View Post
    because the question askes me to find x^n - 1 for n = 6, 7, 8, ..., 22 so I think I have to do for each n.
    Well you've been shown the general way to do it, so write them all I guess.
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  5. #5
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    alright, thank you for helping.
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  6. #6
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    Quote Originally Posted by redsoxfan325 View Post
    x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1}).

    So x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5).

    I don't know if that's what you're looking for.
    But notice that if the exponent is not prime then some further factorisation may be possible. For example, x^6-1 = (x^3)^2-1 = (x^3+1)(x^3-1) = (x+1)(x^2-x+1)(x-1)(x^2+x+1).

    [I'm assuming that you are looking for factorisations where the coefficients are rational numbers. If you're working over the complex numbers then you can factorise x^n-1 into n linear factors. If you are doing it over the real numbers then you can get a factorisation into linear and quadratic factors (but that is just as hard as doing it over the complex numbers).]
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