# Factorization of the polynomial

• August 1st 2011, 08:26 AM
Aero763
Factorization of the polynomial
I need to factor the next polynomial:

$\frac{\sqrt{(2p+1)^3}+\sqrt{(2p-1)^3}}}{\sqrt{4p+2\sqrt{4p^2-1}}}$
• August 1st 2011, 10:29 AM
Opalg
Re: Factorization of the polynomial
Quote:

Originally Posted by Aero763
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.
• August 1st 2011, 10:51 AM
Krizalid
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.
• August 1st 2011, 09:12 PM
Aero763
Re: Factorization of the polynomial
Quote:

Originally Posted by Krizalid
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
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:
$x^2-xy+y^2$

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

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?
• August 1st 2011, 09:19 PM
Krizalid
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.
• August 1st 2011, 09:27 PM
Aero763
Re: Factorization of the polynomial
You are right, I can see it now!