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Math Help - factorizing a 4th degree polynomial

  1. #1
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    factorizing a 4th degree polynomial

    hey guys this is probably pretty basic stuff for most of you. im however struggling with this.
    i don't have any standard procedure for this and i can't figure out how its done. i just don't know what kind of method i should be searching for.
    in my course it was said you are already supposed to know how to do this so im lacking skill on this. here is the problem:

    i underlined the equation where the polynomial is factorized. how to do this? whats the method called and how can i learn to do this on any generic 4th degree polynomial?
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  2. #2
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    Re: factorizing a 4th degree polynomial

    Hey vermillion.

    There is a result that allows you to factor general fourth degree polynomial (which you can assume to be true for algebraic problems):

    Quartic function - Wikipedia, the free encyclopedia
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    Re: factorizing a 4th degree polynomial

    Quote Originally Posted by chiro View Post
    Hey vermillion.

    There is a result that allows you to factor general fourth degree polynomial (which you can assume to be true for algebraic problems):

    Quartic function - Wikipedia, the free encyclopedia
    thanks for your response. but i am afraid this article doesn't really help me here. in the article there is a different last term, an e. I bet there is an easier method out there as well. i hope someone can give another method.

    and also in this case the a is always 1. i bet there is some kind of fancy polynomial trick for this.
    Last edited by vermillion; August 11th 2013 at 04:44 PM.
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    Re: factorizing a 4th degree polynomial

    also, chiro, can you maybe if you think that wikipedia article helps, show me how these 2 examples are factorized? i can't figure it out.
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    Re: factorizing a 4th degree polynomial

    Quote Originally Posted by vermillion View Post
    also, chiro, can you maybe if you think that wikipedia article helps, show me how these 2 examples are factorized? i can't figure it out.
    The {x^3} + 4{x^2} + 6x + 4 is factorized by finding the roots of {x^3} + 4{x^2} + 6x + 4 = 0
    If it has any integer root it must be 1 or 2 or 4 (actually we can skip the positive solutions since we will have all positive members on the equation so they can't give zero)
    Here we see that -2 works and is a root.
    Then dividing {x^3} + 4{x^2} + 6x + 4 with x+2 we get {x^2} + 2{x} + 2 which has roots -1-i and -1+i.
    Ergo its factorization is (x+2)(x+1+i)(x+1-i)

    You can use the same method with the other equation also.
    It's not a general method. I.e it will not work in most cases. For that you have to use the link chiro gave.
    Last edited by ChessTal; August 12th 2013 at 01:34 PM.
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    Re: factorizing a 4th degree polynomial

    Quote Originally Posted by vermillion View Post
    thanks for your response. but i am afraid this article doesn't really help me here. in the article there is a different last term, an e. I bet there is an easier method out there as well. i hope someone can give another method.

    and also in this case the a is always 1. i bet there is some kind of fancy polynomial trick for this.
    Well there is a "fancy" method for a=1:
    The factorization is (x - {r_1})(x - {r_2})(x - {r_3})(x - {r_4}) where {r_1},{r_2},{r_3},{r_4} can be found from here: quartic formula | planetmath.org
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    Re: factorizing a 4th degree polynomial

    very fancy indeed ok thanks guys. However i found out i was overcomplicating as apparently in the exams there won't be any cases where you have to solve 4th degree polynomials, in the worst case they are 3th and i can do that. Also i found a note in the lectures that states 4th degrees should be solved numerically but we won't be using calculators during exams so im good.
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  8. #8
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    Re: factorizing a 4th degree polynomial

    Also you might want to consider the guessing method were you guess a root and then use long division to go from say cubic to quadratic (in which case you use the quadratic formula or further guesses).

    So for example if I had say (x-1)(x-2)^2
    = (x-1)(x^2 - 4x + 4)
    = (x^3 - 4x^2 + 4x - x^2 + 4x - 4)
    = x^3 - 5x^2 + 8x - 4

    If you tried a test solution of x = 1 you would get zero and thus you could factor (x-1) straight out.
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