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Math Help - Factorisation

  1. #1
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    Factorisation

    Factorize a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)
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  2. #2
    Senior Member Dinkydoe's Avatar
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    It seems like some obvious combination like (\pm a^2\pm b^2\pm c)^2 doesn't work for any combination of +/-

    We can write it as a polynomial g with:

    c=x

    and

    g(x)= x^2-2(a^2+b^2)x+(a^4+b^4-2a^2b^2)

    and then use abc-formula to find roots of g(x)=0
    Then we can write (x-r_1)(x-r_2) as a factorisation. (r_1,r_2 roots)

    But such factorisation is not quite unique, in this case...
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  3. #3
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    Quote Originally Posted by vuze88 View Post
    Factorize a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)
    Dear vuze88,

    Notice that, (a^2+b^2-c)^2=a^4+b^4+c^2+2a^2b^2-2a^2c-2b^2c

    Therefore, (a^2+b^2-c)^2-4a^2b^2=a^4+b^4+c^2-2a^2b^2-2a^2c-2b^2c

    a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=(a^2+b^2-c)^2-4a^2b^2=(a^2+b^2-c-2ab)(a^2+b^2-c+2ab)

    a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=((a-b)^2-c)((a+b)^2-c)

    Hope this helps.
    Last edited by Sudharaka; June 22nd 2010 at 03:19 AM.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Sudharaka View Post
    Dear vuze88,

    Notice that, (a^2+b^2-c)^2=a^4+b^4+c^2+2a^2b^2-2a^2c-2b^2c

    Therefore, (a^2+b^2-c)^2-4a^2b^2=a^4+b^4+c^2-2a^2b^2-2a^2c-2b^2c

    a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=(a^2+b^2-c)^2-4a^2b^2=(a^2+b^2-c-2ab)(a^2+b^2-c+2ab)

    a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=((a-b)^2-c)((a+b)^2+c)

    Hope this helps.
    Really nice solution, just thought I should point out a small typo; the last "+" sign should be a "-".

    Cheers.
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  5. #5
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    Quote Originally Posted by undefined View Post
    Really nice solution, just thought I should point out a small typo; the last "+" sign should be a "-".

    Cheers.
    Dear undefined,

    Thanks. Corrected it.
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