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  1. MHB Journeyman
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    Fernando Revilla's Avatar
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    #11
    Quote Originally Posted by bwpbruce View Post
    Which of my expressions are you implying is something other than closed form?
    (1) You write $$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$$ What happens when you substitute $b=\pm\dfrac{1}{2}$?

    (2) The title of your thread Factoring polynomial, usually means to express a polynomial as a product of irreducible polynomial and does not include rational fractions. Otherwise, you should specify the terms of the factorization you are referring to.

  2. MHB Apprentice

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    #12
    Quote Originally Posted by Fernando Revilla View Post
    (1) You write $$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$$ What happens when you substitute $b=\pm\dfrac{1}{2}$?

    (2) The title of your thread Factoring polynomial, usually means to express a polynomial as a product of irreducible polynomial and does not include rational fractions. Otherwise, you should specify the terms of the factorization you are referring to.
    You're still asking about $b=\pm\dfrac{1}{2}$?.

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