radical factorization

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March 11th 2006, 06:15 PM
guess

radical factorization

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

Thanks Bro
March 11th 2006, 06:26 PM
ThePerfectHacker

Quote:

Originally Posted by guess

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

Thanks Bro

You mean decomposition of fractions?
Such as,
$\frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$
March 12th 2006, 01:59 AM
guess

I mean something more complication such as 1/(3+X^5)
March 12th 2006, 02:01 AM
guess

Or more complex like 1/(1+x^6)
March 12th 2006, 07:21 AM
Aradesh

I think you are referring to your integration thread when $1+x^4$ was factorized to $(x^2 + \sqrt 2 x + 1)(x^2 -\sqrt2 x + 1)$ by TD.

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

now how would you think to do this?
well it comes from the difference of two squares where if you have $x^2 - y^2$ this is $(x+y)(x-y)$

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
$( (a+b) + \alpha )( (a+b) - \alpha )$
this expands to
$a^2 + b^2 + 2ab - \alpha^2$
so if we choose $\alpha$ such that:
$\alpha^2 = 2ab$
then:
$( (a+b) + \alpha )( (a+b) - \alpha ) = a^2 + b^2$
so
$\alpha = \sqrt{2ab}$
we get:
$(a+b+\sqrt{2ab})(a+b-\sqrt{2ab})$
so from the original $1+x^4$
if we say a = 1 and b = x^2 it factorizes to:
$(x^2 + \sqrt 2 x + 1)(x^2 -\sqrt2 x + 1)$
March 12th 2006, 07:27 AM
Aradesh

so in your integration thread you had
$\frac{1}{x^4+1}$
and as shown the bottom part factorizes like so
$<br /> \frac{1}{{x^4 + 1}} = \frac{1}{{\left( {x^2 + \sqrt 2 x + 1} \right)\left( {x^2 - \sqrt 2 x + 1} \right)}}<br />$
then this was then decomposed with partial fractions to the form (i'll take td's word):
$<br /> \frac{1}{{x^4 + 1}} = \frac{{\sqrt 2 \left( {\sqrt 2 - x} \right)}}{{4\left( {x^2 - \sqrt 2 x + 1} \right)}} + \frac{{\sqrt 2 \left( {\sqrt 2 + x} \right)}}{{4\left( {x^2 + \sqrt 2 x + 1} \right)}}<br />$

which is what theperfecthacker was talking about.
March 12th 2006, 07:44 AM
Aradesh

it is worth noting however that
$1+x^4$ is difficult to factorize because it is in the form
$x^{2^n}+1$
factorizing things in the form $x^k+1$ for all odd k
$x^k+1 = (x+1)(x^{k-1} - x^{k-2} + ... - x + 1)$
so $x^p + 1$ 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:
$(x^k+1) = (x^{rp} + 1) = ((x^r)^p + 1)$ and since p is odd, this factorizes also, to:
$((x^r) + 1)((x^r)^{p-1} - (x^r)^{p-2} + ... - x^r + 1)$
so this only leaves things in the form
$x^{2^n} + 1$ which don't factorize nicely.
$z^2 + 1$ does not factorize
if $z = x^{2^{n-1}}$
then
$z^2 + 1 = x^{2^n}+1$
so $x^{2^n} + 1$ does not factorize either.
this is useful in searching for fermat primes, so we only know to look for primes in the form $2^{2^n} + 1$ since all other indices factorize.
March 13th 2006, 01:30 AM
guess

thanks alot aradesh... You have done a Great help...