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Math Help - Factorization of the polynomial

  1. #1
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    Factorization of the polynomial

    I need to factor the next polynomial:




    \frac{\sqrt{(2p+1)^3}+\sqrt{(2p-1)^3}}}{\sqrt{4p+2\sqrt{4p^2-1}}}
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  2. #2
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    Re: Factorization of the polynomial

    Quote Originally Posted by Aero763 View Post
    I need to factor the next polynomial:

    \frac{\sqrt{(2p+1)^3}+\sqrt{(2p-1)^3}}}{\sqrt{4p+2\sqrt{4p^2-1}}}
    For a start, it's not a polynomial. But it is possible to simplify it. As a hint, let x = \sqrt{2p+1} and y = \sqrt{2p+1}, and express everything in terms of x and y.
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  3. #3
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    Re: Factorization of the polynomial

    Since \sqrt{{{\left( \sqrt{2p+1}+\sqrt{2p-1} \right)}^{2}}}=\sqrt{2p+1}+\sqrt{2p-1}, which is the denominator, you can factor the numerator as a sum of perfect cubes.
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    Re: Factorization of the polynomial

    Quote Originally Posted by Krizalid View Post
    Since \sqrt{{{\left( \sqrt{2p+1}+\sqrt{2p-1} \right)}^{2}}}=\sqrt{2p+1}+\sqrt{2p-1}, which is the denominator, you can factor the numerator as a sum of perfect cubes.
    Thank you, but I already tried that to no avail.

    Quote Originally Posted by Opalg View Post
    For a start, it's not a polynomial. But it is possible to simplify it. As a hint, let x = \sqrt{2p+1} and y = \sqrt{2p+1}, and express everything in terms of x and y.
    Thank you, that worked like a charm! I already came up with the idea, but was not sure what to do with 4p in the denominator. But since you suggested it, I took a closer look and sure enough - it worked!


    There how it goes:

    1) First we need to do something with 4p, which confused me, when I tried to solve it by myself. We need to get it in the form of (2p+1) and (2p-1) somehow. Well, 4p is equal to 2p + 2p, and we can add and substract 1. Therefore:
    4p = 2p - 1 + 2p + 1


    Here we go


    \frac{\sqrt{(2p+1)^3}+\sqrt{(2p-1)^3}}}{\sqrt{2p - 1 + 2p + 1 +2\sqrt{4p^2-1}}}

    Now, let x = \sqrt{2p+1} and y = \sqrt{2p-1}, and express everything in terms of x and y:


    \frac{x^3+y^3}{\sqrt{x^2 + y^2 + 2xy}}



    By definition x^2 + y^2 + 2xy = (x+y)^2, and x^3+y^3=(x+y)(x^2-xy+y^2), therefore:
    \frac{(x+y)(x^2-xy+y^2)}{(x+y)}

    Simplify:
    x^2-xy+y^2


    And the answer is:

    4p - \sqrt{4p^2-1}


    Thank you guys again, been trying to solve this tough cookie for a couple of days now.


    P.S. if it is not a polynomial, then what is it?
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  5. #5
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    Re: Factorization of the polynomial

    Opalg's hint is just to make things easier to work, haha, you coulda solved the problem directly, with no substitutions.
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  6. #6
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    Re: Factorization of the polynomial

    You are right, I can see it now!
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