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Math Help - Factorisation

  1. #1
    M.R
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    Factorisation

    Hi,

    I have managed to factorise A^4 - B^4. But can't factorise A^4 + B^4. Can anyone help me out? Thanks
    Last edited by M.R; December 24th 2009 at 04:52 PM.
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  2. #2
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    Quote Originally Posted by M.R View Post
    Hi,

    I have managed to factorise A^4 - B^4. But can't factorise A^4 + B^4. Can anyone help me out? Thanks
    You can't without using complex numbers, have you met these?
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  3. #3
    Super Member Bacterius's Avatar
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    Since any expression a^{2n} with n \in \mathbb{N} is always positive, then a^{2n} + b^{2n'} will always be positive. Thus, there is no possible factorization of such expressions in \mathbb{R}. However, there is a factorization in \mathbb{C}, which involves complex numbers.
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  4. #4
    M.R
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    I haven't done. We just started the maths course. In Chapter 1 we have so far done:
    1A - Terms, Factors and indices
    1B - Expanding Brackets
    1C - Factorisation
    1D - Algebraic Factions
    1E - Four Cubic Identities. And this is where we are up to.

    The answer for A^4 + B^4 is given as (A^2 - \sqrt{2}AB + B^2) (A^2 + \sqrt{2}AB + B^2).

    So i dont' think they used complex numbers. I have tired expanding out the answer and it is correct. But i can't figure it out, how to get there?
    Last edited by M.R; December 24th 2009 at 04:51 PM.
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  5. #5
    Super Member Bacterius's Avatar
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    No, it is wrong. You cannot obtain a full factorization of a^4 + b^4 in \mathbb{R}.

    Here is an example : someone asks you to find the solutions of 6^4 + x^4 = 0. You would apply your factorization to this problem, so it becomes :

    (6^2 - \sqrt{2} \times 6x + x^2) (6^2 + \sqrt{2}\times 6x + x^2) = 0

    Thus we are left with :

    x^2 - \sqrt{2}\times 6x + 36 = 0

    and

    x^2 + \sqrt{2}\times 6x + 36 = 0

    Oops ! None of these quadratic equations admit roots in \mathbb{R}. Now you are screwed without complex numbers.

    ----------------

    I rather think the factorization they come up with is a bit useless, since any use you would eventually make of this factorization would anyway require complex numbers to be successfully accomplished.
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  6. #6
    M.R
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    Hi Bacterius,

    The question is from Cambridge Mathematics 3 Unit (year 11). Exercise 1E, question 10.

    Furthermore, similar answer are given from question 11. Please have a look at http://www.mathhelpforum.com/math-he...orisation.html

    So i don't know how else to figure it out with basic maths. Mind you it is from the extension question section
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  7. #7
    Super Member Bacterius's Avatar
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    I know what you mean.
    The factorization you are given is correct, but you will need to learn complex numbers to effectively use it to solve problems (furthermore, you will learn a much simpler factorization involving complex numbers when you will have learnt the later).
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  8. #8
    M.R
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    Yes, your right. I am still a bit far from complex number. I think it's covered in the 4 unit (extension 2) course. But thanks again for you help
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  9. #9
    M.R
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    Thanks to Soroban, from post (http://www.mathhelpforum.com/math-he...orisation.html) i have figured it out:

    A^4+B^4
    =A^4+2A^2B^2+B^4 - 2A^2B^2
    =(A^2+B^2)^2 - (\sqrt{2}AB)^2
    =(A^2-\sqrt{2}AB+B^2) (A^2+\sqrt{2}AB+B^2) [Difference of squares]
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