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Math Help - Factorisation

  1. #1
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    Factorisation

    Hi.

    How would go about factoring x^6+1

    Thanks
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  2. #2
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    what about (x^3+i)(x^3-i)

    expanding back out to check

    x^6+x^3i-x^3i-i^2

    x^6-i^2

    x^6-(-1)

    x^6+1
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  3. #3
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    Quote Originally Posted by deltaxray View Post
    Hi.

    How would go about factoring x^6+1

    Thanks
    If you're factorising over the Real Numbers, it does not factorise.

    If you're factorising over the Complex Numbers, change it to a difference of two squares.

    x^6 + 1 = (x^3)^2 - i^2

     = (x^3 + i)(x^3 - i).


    You may be able to use a sum or difference of two cubes factorisation now as well...
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  4. #4
    Senior Member Stroodle's Avatar
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    You can factorise it to:

     ( x^2+\sqrt{3}x+1 ) ( x^2-\sqrt{3}x+1 ) ( x^2+1 )

    But unfortunately I cant remember the rule for doing this
    Last edited by Stroodle; November 30th 2009 at 08:17 AM. Reason: fixed error
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  5. #5
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    Quote Originally Posted by Stroodle View Post
    You can factorise it to:

     ( x^2+\sqrt{3}x+1 ) ( x^2-\sqrt{3}x+1 ) ( x+1 )

    But unfortunately I cant remember the rule for doing this
    You have a typo. The (x + 1) should be (x^2 + 1). See below.

    Quote Originally Posted by deltaxray View Post
    Hi.

    How would go about factoring x^6+1

    Thanks
    x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)((x^2)^2 - x^2 + 1^2)

    using the sum of two cubes formula

     = (x^2 + 1)(x^4 - x^2 + 1).

    Now note that x^4 - x^2 + 1 = x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - 3x^2 = (x^2 + 1 - \sqrt{3} x) (x^2 + 1 + \sqrt{3} x) using the difference of two squares formula.
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