Can someone teach me how to do radical factorization? Or introduce a good website which teaches how to do it..

Thanks Bro

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- Mar 11th 2006, 06:15 PMguessradical factorization
Can someone teach me how to do radical factorization? Or introduce a good website which teaches how to do it..

Thanks Bro - Mar 11th 2006, 06:26 PMThePerfectHackerQuote:

Originally Posted by**guess**

Such as,

$\displaystyle \frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$ - Mar 12th 2006, 01:59 AMguess
I mean something more complication such as 1/(3+X^5)

- Mar 12th 2006, 02:01 AMguess
Or more complex like 1/(1+x^6)

- Mar 12th 2006, 07:21 AMAradesh
I think you are referring to your integration thread when $\displaystyle 1+x^4$ was factorized to $\displaystyle (x^2 + \sqrt 2 x + 1)(x^2 -\sqrt2 x + 1)$ by TD.

you can do this since if you have $\displaystyle a^2 + b^2$

it can be expressed in the form $\displaystyle (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})$ if you expand the brackets you will see that it is indeed $\displaystyle a^2 + b^2$

now how would you think to do this?

well it comes from the difference of two squares where if you have $\displaystyle x^2 - y^2$ this is $\displaystyle (x+y)(x-y)$

if we think of our x in the difference of two squares as (a+b) then (a+b)^2 is going to come out in the result somewhere, and (a+b)^2 = a^2 + b^2 + 2ab

now this has a^2+b^2 in it but it is 2ab too much.

however if we choose the right y we can get the 2ab to disappear. for instance

$\displaystyle ( (a+b) + \alpha )( (a+b) - \alpha )$

this expands to

$\displaystyle a^2 + b^2 + 2ab - \alpha^2$

so if we choose $\displaystyle \alpha$ such that:

$\displaystyle \alpha^2 = 2ab $

then:

$\displaystyle ( (a+b) + \alpha )( (a+b) - \alpha ) = a^2 + b^2$

so

$\displaystyle \alpha = \sqrt{2ab}$

we get:

$\displaystyle (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})$

so from the original $\displaystyle 1+x^4$

if we say a = 1 and b = x^2 it factorizes to:

$\displaystyle (x^2 + \sqrt 2 x + 1)(x^2 -\sqrt2 x + 1)$ - Mar 12th 2006, 07:27 AMAradesh
so in your integration thread you had

$\displaystyle \frac{1}{x^4+1}$

and as shown the bottom part factorizes like so

$\displaystyle

\frac{1}{{x^4 + 1}} = \frac{1}{{\left( {x^2 + \sqrt 2 x + 1} \right)\left( {x^2 - \sqrt 2 x + 1} \right)}}

$

then this was then decomposed with partial fractions to the form (i'll take td's word):

$\displaystyle

\frac{1}{{x^4 + 1}} = \frac{{\sqrt 2 \left( {\sqrt 2 - x} \right)}}{{4\left( {x^2 - \sqrt 2 x + 1} \right)}} + \frac{{\sqrt 2 \left( {\sqrt 2 + x} \right)}}{{4\left( {x^2 + \sqrt 2 x + 1} \right)}}

$

which is what theperfecthacker was talking about. - Mar 12th 2006, 07:44 AMAradesh
it is worth noting however that

$\displaystyle 1+x^4$ is difficult to factorize because it is in the form

$\displaystyle x^{2^n}+1$

factorizing things in the form $\displaystyle x^k+1$ for all odd k

$\displaystyle x^k+1 = (x+1)(x^{k-1} - x^{k-2} + ... - x + 1)$

so $\displaystyle x^p + 1$ where p is a prime number not equal to 2 is factorizable into the form above. as the rest of the prime numbers are odd.

so for even k, if k = rp where p is a prime factor other than 2 then:

$\displaystyle (x^k+1) = (x^{rp} + 1) = ((x^r)^p + 1)$ and since p is odd, this factorizes also, to:

$\displaystyle ((x^r) + 1)((x^r)^{p-1} - (x^r)^{p-2} + ... - x^r + 1)$

so this only leaves things in the form

$\displaystyle x^{2^n} + 1$ which don't factorize nicely.

$\displaystyle z^2 + 1$ does not factorize

if $\displaystyle z = x^{2^{n-1}}$

then

$\displaystyle z^2 + 1 = x^{2^n}+1$

so $\displaystyle x^{2^n} + 1$ does not factorize either.

this is useful in searching for fermat primes, so we only know to look for primes in the form $\displaystyle 2^{2^n} + 1$ since all other indices factorize. - Mar 13th 2006, 01:30 AMguess
thanks alot aradesh... You have done a Great help...