I need to factor the next polynomial:

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

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- Aug 1st 2011, 08:26 AMAero763Factorization of the polynomial
I need to factor the next polynomial:

$\displaystyle \frac{\sqrt{(2p+1)^3}+\sqrt{(2p-1)^3}}}{\sqrt{4p+2\sqrt{4p^2-1}}}$ - Aug 1st 2011, 10:29 AMOpalgRe: Factorization of the polynomial
- Aug 1st 2011, 10:51 AMKrizalidRe: Factorization of the polynomial
Since $\displaystyle \sqrt{{{\left( \sqrt{2p+1}+\sqrt{2p-1} \right)}^{2}}}=\sqrt{2p+1}+\sqrt{2p-1},$ which is the denominator, you can factor the numerator as a sum of perfect cubes.

- Aug 1st 2011, 09:12 PMAero763Re: Factorization of the polynomial
Thank you, but I already tried that to no avail.

Thank you, that worked like a charm! I already came up with the idea, but was not sure what to do with $\displaystyle 4p$ 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 $\displaystyle 4p$, which confused me, when I tried to solve it by myself. We need to get it in the form of $\displaystyle (2p+1)$ and $\displaystyle (2p-1) $ somehow. Well, $\displaystyle 4p$ is equal to $\displaystyle 2p + 2p$, and we can add and substract 1. Therefore:

$\displaystyle 4p = 2p - 1 + 2p + 1$

Here we go

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

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

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

By definition $\displaystyle x^2 + y^2 + 2xy = (x+y)^2$, and $\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$, therefore:

$\displaystyle \frac{(x+y)(x^2-xy+y^2)}{(x+y)}$

Simplify:

$\displaystyle x^2-xy+y^2$

And the answer is:

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

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? - Aug 1st 2011, 09:19 PMKrizalidRe: 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.

- Aug 1st 2011, 09:27 PMAero763Re: Factorization of the polynomial
You are right, I can see it now!