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Math Help - Complicated factorisation technique

  1. #1
    Newbie
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    Complicated factorisation technique

    Hi,

    Does anyone know a reasonably standard technique to factorise something like,

    x^4 + y^4 + 9x^2 + y^2 - 6x^3 + 2x^2y^2 - 6xy^2 - 4x = 0

    into:

    (x^2-4x+y^2) (x^2-2x+y^2+1) = 0

    i.e. when not given any other information (like one of the factors), how to recognise it can be factored, and do it etc
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  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
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    BeiJing
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    notice that the degree of y is 2 or 4, we can let t=y^2,\text{ then }t^2=y^4.
    consider the left hand side as a quadratical polynomial of t , we can rewrite it as
    t^2+(2x^2-6x+1)t+(x^4-6x^3+9x^2-4x)=0.
    then factorise (x^4-6x^3+9x^2-4x):
    since the root of (x^4-6x^3+9x^2-4x) is 0 , 1(multiple root) , 4, we have (x^4-6x^3+9x^2-4x)=x(x-4)(x-1)^2.
    then we find the fact that (2x^2-6x+1) is exactly the sum of x(x-4)\text{ and }(x-1)^2.
    therefore the factorization follows immediately.

    I hope I've made myself clear enough, and can help you out. Good luck!
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