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Math Help - Factoring a large Diff of Squares

  1. #1
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    Red face Factoring a large Diff of Squares

    y = x^1000 - 1

    I would like to factor out an x-1 from this but I don't remember which rules apply.

    1) Do I apply something similar to a difference of squares and just keep going until I get (x-1)(....)?

    (x^500)^2 - (1)^2 = (x^500+1)(x^500-1) = (x^500+1)(x^250+1)(x^250-1) = (x^500+1)(x^250+1)(x^125+1)(x^125-1) = now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent ... well I'll try difference of cubes in that last factor...: = (x^500+1)(x^250+1)(x^125+1)(x^5-1)(x^10+x^5+1)... Okay, now I'm stuck, unless I remember how to factor (x^10+x^5+1)...


    2) Or do I use some kind of Binomial Theorem.

    Is it true that
    x^1000 - 1 = (x-1)(x^999+x^998+x^997......+1) ?


    3) Do I have to remember Pascal's Triangle at all?

    For example, if number 2 above is true, then do I have to put in pascal's coefficients? ie.
    (x-1)(1x^999 + 999x^998 + ...... + 1)?
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  2. #2
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    Hello, krystaline86!

    This is an ugly problem . . .


    Factor: . y \:= \:x^{1000} - 1

    1) Do I apply something similar to a difference of squares
    and just keep going until I get: (x-1)(\cdots)?

    (x^{500})^2 - (1)^2 \;= \;(x^{500}+1)(x^{500}-1)

    . . . =\;(x^{500}+1)(x^{250}+1)(x^{250}-1)

    . . . = \;(x^{500}+1)(x^{250}+1)(x^{125}+1)(x^{125}-1) . . . . Good!

    now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent.

    But odd exponents are usually "a good thing".
    From what you wrote, I assume you are familiar with these formulas:

    . . \begin{array}{ccc}a^5 - 1 &=&(a-1)(a^4+a^3+a^2+a+1)\\ \\[-4mm]<br />
a^5 + 1 &=&(a+1)(a^4-a^3+a^2-a+1) \end{array}


    So: . \begin{array}{ccccccc}x^{125}-1 &=& \left(x^{25}\right)^5 - 1 &=& \left(x^{25} - 1\right)\left(x^{100} + x^{75} + x^{50} + x^{25} + 1\right)  \\<br />
x^{125} + 1 &=& \left(x^{25}\right)^5 + 1 &=& \left(x^{25} + 1\right)\left(x^{100} - x^{75} + x^{50} - x^{25} + 1\right) \end{array}


    Then: . \begin{array}{ccccccc}x^{25}-1 &=& \left(x^5\right)^5-1 &=& (x^5-1)(x^{20} + x^{15} + x^{10} + x^5+1) \\<br />
x^{25} + 1 &=&\left(x^5\right)^5+1 &=& (x^5+1)(x^{20} - x^{15} + x^{10} - x^5 + 1) \end{array}


    And: . \begin{array}{ccccc}x^5 - 1 &=&(x-1)(x^4 + x^3 + x^2 + x+1) \\ x^5+1 &=& (x+1)(x^4 - x^3 + x^2 - x + 1) \end{array}


    . . See?

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  3. #3
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    Excellent. So my idea about using the Binomial Theorem was right then?

    The answer would be:

    (x^1000 - 1) = (x-1)(x^999+x^998+x^997+...+x+1)

    Right?
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  4. #4
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    I just want to know what that second factor expanded out would look like in the form of how I have it there.

    OH, and how do I express that properly? Can I use a factorial somehow?
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