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Math Help - Polynomials - Solving Equations

  1. #1
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    Cool Polynomials - Solving Equations

    I've done quite a lot of practice with this topic but I'm stuck at these kind of questions.

    Q) Find all the real roots of the equation x^5-3x^4+2x^3-2x^2+3x+1=2x

    I've solved one root which is 1, but I got that by using the guess factor and putting in different values of x to equate the equation.

    Can you show me the easiest way of solving these kind of questions.

    Thanks in advance!
    Last edited by unstopabl3; July 16th 2010 at 02:13 PM.
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  2. #2
    A Plied Mathematician
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    Quintics are dicey. In fact, there is a proof that although there are formulas for the general quadratic, cubic, and quartic polynomials, there is no general formula that solves the quintic. You can only solve special cases. In this case, 1 and -1 would be the numbers to try, because of the rational root theorem. After trying those, if you can reduce the order by factoring, you might be able to use the cubic formula or the quartic method.
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  3. #3
    Senior Member eumyang's Avatar
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    Ackbeet mentioned the rational roots theorem, which gives us 1 and -1 as possibilities. Divide the original polynomial by x - 1 using synthetic division and see what you get.
    Last edited by mr fantastic; July 16th 2010 at 02:37 PM. Reason: OP fixed typo.
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  4. #4
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    Cool

    Well this is a High School level question, so I am assuming there must be a simpler method to solve this? How can I factorize this?

    Better yet can someone show me an easy method of solving such questions.

    Thanks!
    Last edited by mr fantastic; July 16th 2010 at 02:37 PM.
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  5. #5
    Super Member bigwave's Avatar
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    Cool

    x^5-3x^4+2x^3-2x^2+x+1=0

    real roots are 1, 1\pm\sqrt{2}
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  6. #6
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    Well that is correct, but how can I solve this myself. Can you please show me the method.
    Thanks!
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  7. #7
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    Quote Originally Posted by unstopabl3 View Post
    Well that is correct, but how can I solve this myself. Can you please show me the method.
    Thanks!
    Ackbeet's post told you how to find the real root x = 1.

    x = 1 is a root => x - 1 is a factor. Divide that factor into your quintic using polynomial long division and you get x^4 - 2x^3 -2x-1 as another factor.

    Your new job now is to solve x^4 - 2x^3 -2x-1 = 0.

    Note that the left hand side of the above can be written x^4 - 2x^3 - x^2 + x^2 - 2x - 1 = x^2(x^2 - 2x - 1) + ( x^2 - 2x - 1) ....
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  8. #8
    Super Member bigwave's Avatar
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    well what is left from the first factor is
    (x^2+1)(x^2-2x-1)
    to find factors of (x^2-2x-1)  \rightarrow (x-1)^2 -2  \rightarrow x= 1\pm\sqrt{2}
    this was done by completing the square.

    do you need more steps.
    i don't think there is a slam dunk to this one.

    EDIT: clarified step
    Last edited by bigwave; July 16th 2010 at 03:04 PM. Reason: claification
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  9. #9
    Senior Member eumyang's Avatar
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    x^5 - 3x^4 + 2x^3 - 2x^2 + x + 1=0
    As I said, since you know that one of the roots is 1, use synthetic division:
    Code:
    1|  1 -3  2 -2  1  1
    --     1 -2  0 -2 -1
    ---------------------
        1 -2  0 -2 -1  0
    So your equation is the same as
    (x - 1)(x^4 - 2x^3 - 2x - 1) = 0

    Take the 2nd factor and solve for x:
    x^4 - 2x^3 - 2x - 1 = 0

    Subtract and add an x^2 on the left side:
    x^4 - 2x^3 - x^2 + x^2 - 2x - 1 = 0

    Factor by grouping three terms -- find the GCF for the first 3 terms, and do the same for the last 3 terms:
    x^2(x^2 - 2x - 1) + 1(x^2 - 2x - 1) = 0

    So this factors to
    (x^2 + 1)(x^2 - 2x - 1) = 0

    Set each factor to 0. For the first, there will be no real solutions. For the 2nd, use the quadratic formula.

    EDIT: Ack! Too slow!
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