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Math Help - Factorizing cubics

  1. #1
    Member Jones's Avatar
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    Factorizing cubics

    Hi,

    Is there some kind of formula for factorizing cubic s?

    By formula i mean. If we have a second degree polynomial x^2+6x+12 we know we need two numbers that multiply to give us 12 and add together to give us 6. Is there any similar forumula for cubics?

    Im having some difficulties factorizing the expression  x^4-11x^3-36x^2-36 i know that 2 is a factor so after taking out x-2 i end up with x^3-9x^2+18

    The closest i got was (x-6x)(x^2-3x) doesn't really cut it though =/
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  2. #2
    Eater of Worlds
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    Actually, for that quartic, 2 is not a factor.

    It has 2 real and 2 complex solutions. The real ones are not nice.

    You do mean x^{4}-11x^{3}-36x^{2}-36

    Make sure you did not make a typo with a sign.
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  3. #3
    Member Jones's Avatar
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    Quote Originally Posted by galactus View Post
    Actually, for that quartic, 2 is not a factor.

    It has 2 real and 2 complex solutions. The real ones are not nice.

    You do mean x^{4}-11x^{3}-36x^{2}-36

    Make sure you did not make a typo with a sign.


    Hmm, you're right, i must have missed(or added) a parenthesis somewhere =/

    But anyway is there an easy way of solving these equations?
    i know the usual trick won't work here since we have an x^3 term

    Otherwise you could set x^2=t
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Hmm, you're right, i must have missed(or added) a parenthesis somewhere =/

    But anyway is there an easy way of solving these equations?
    i know the usual trick won't work here since we have an x^3 term

    Otherwise you could set x^2=t
    If a polynomial with integer coefficients has "nice" factors (that is the polynomial equation that you get by setting the polynomial to zero has rational roots) the rational roots theorem tell us they are the ratio of a factor of the constant term to a factor of the coefficient of the highest power of the variable.

    That means for your cubic:

    <br />
x^{4}-11x^{3}-36x^{2}-36<br />

    the candidates for rational roots are \pm 1,\ \pm 2,\ \pm 3,\ \pm 6,\ \pm 9,\,\ \pm 12,\  \pm 18,\ \pm 36 we try all of these and find that none of them is a root, so this quartic has no nice factors.

    RonL
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  5. #5
    Member Jones's Avatar
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    So, um.. how would you solve something like that?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jones View Post
    So, um.. how would you solve something like that?
    Graphically or numerically. The point is that your question is flawed somehow.

    -Dan
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    So, um.. how would you solve something like that?
    Depends; numerical and graphical solutions are often adequate, but the is an algebraic method which for most practical purposes is useless, but see here.

    RonL
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