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Math Help - radical factorization

  1. #1
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    radical factorization

    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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  2. #2
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    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}
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  3. #3
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    I mean something more complication such as 1/(3+X^5)
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  4. #4
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    Or more complex like 1/(1+x^6)
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  5. #5
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    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)
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  6. #6
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    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.
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  7. #7
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    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.
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  8. #8
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    thanks alot aradesh... You have done a Great help...
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