1. Originally Posted by bwpbruce
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. Originally Posted by Fernando Revilla
(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 $b=\pm\dfrac{1}{2}$?.