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Math Help - Factorising Polynomial over the real number field

  1. #1
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    Factorising Polynomial over the real number field

    hey guys I was just wondering whether I'm doing this right
    for example the question asks to factor z^6+8 over the real number field

    Heres what I did
    z^6+8=(z^2)^3+2^3
    =(z^2+2)(z^4-2z^2+4)

    but when I checked the answers they got
    z^6+8=(z^2+2)(z^2+\sqrt6z+2)(z^2-\sqrt6z+2)
    they got this by pairing up the roots

    which is more factored than my expression, the factors I got are not factorised enough right?, I have tried difference of 2 squares, yet they don't give me the same as the answer as they are complex
    =(z^2+2)(z^4-2z^4+4)
    =(z^2+2)((z^2+1)^2-3i^2)
    =(z^2+2)(z^2+1-\sqrt3i)(z^2+1+\sqrt3i)


    • Is the method I used originally wrong, since it is not fully factored?
    • How would I be able to factorise it to the right answer (eg. provided by the textbook) by just continuing with the factors I originally got, without using the pairing up of roots method the text book used. Is it possible?

    Thanks
    Aonin
    Last edited by aonin; May 17th 2011 at 05:48 PM.
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  2. #2
    MHF Contributor
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    Difference of squares? Sure they do. Just rewrite a little:

    z^{4} - 2\cdot z^{2} + 4 = z^{4} + 4 - 2\cdot z^{2} = (z^{2} + 2)^{2} - 2\cdot z^{2} - 4\cdot z^{2} = (z^{2} + 2)^{2} - 6\cdot z^{2}

    You're just not used to completing the square on the usually-outside pieces. Add it to your bag of tricks. You might need it again.
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  3. #3
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    thanks!...never thought about that
    so by completing the square on the outside pieces, its possible to factor any polynomial that couldn't be factored otherwise..or is this a generalisation?
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  4. #4
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    Didn't you ever demonstarte the development of the Quadratic Formula?

    x^{4} + b^{2}x^{2} + c^{2} =

    \left(x^{4} + c^{2}\right) + b^{2}x^{2} =

    (x^{2}+c)^{2} + b^{2}x^{2} - 2cx^{2} =

    (x^{2}+c)^{2} + \left(b^{2} - 2c\right)x^{2} =

    (x^{2}+c)^{2} - \left(2c - b^{2}\right)x^{2} =

    That's an important point. 2c - b^2 > 0 I'm guessing you'll just get more imaginaries otherwise.

    (x^{2} + c + \sqrt{2c - b^{2}}|x|)\cdot (x^{2} + c - \sqrt{2c - b^{2}}|x|)

    I'm going to stick my neck out and say that you may have a chance of finding this result in a book on quartic equations by Orson Pratt in the middle 19th century. It's not my favorite result. I'm not very happy about the absolute values, either.

    I think, if you force a few things into this form, something may occasionally come of it.
    Last edited by TKHunny; May 17th 2011 at 08:02 PM. Reason: Add Absolute Values
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