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Math Help - roots with multiplicity > 1 polynomials

  1. #1
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    Lightbulb roots with multiplicity > 1 polynomials

    I was solving the problem listed here..

    Polynomial x

    I understood the logic behind the first line which says that "A root is of multiplicity 1 ==> the polynomial has a root in common with it's own derivative.

    But I wasn't able to follow the next where the reply is like this.
    Thus the given information tells us that x^4 + x + 6 and 4x^3 + 1 have a common root r. It follows that
    0 = 4(r^4 + r + 6) - r(4r^3 + 1) = 3r + 24.

    I could not follow the logic behind the line 0 = 4(r^4 + r + 6) - r(4r^3 + 1) = 3r + 24.

    I have been searching for the justification behind this step, but have been failing again and again. Can someone throw some light on this?

    Thanks
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: roots with multiplicity > 1 polynomials

    Attached Thumbnails Attached Thumbnails roots with multiplicity > 1 polynomials-f-x-f-x-.png  
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  3. #3
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    Re: roots with multiplicity > 1 polynomials

    if r is a root of x4 + x + 6 and 4x3 + 1, then:

    r4 + r + 6 = 0, and 4r3 + 1 = 0, hence:

    4(r4 + r + 6) - r(4r3 + 1) = 4*0 - r*0 = 0.

    expanding, we have:

    4r4 + 4r + 24 - 4r4 - r = 0, and the r4 terms cancel out leaving:

    3r + 24 = 0

    @MaxJasper:

    the original question deals with a field of characteristic p, so a real-valued plot is not especially helpful, here (since char(R) = 0).
    Last edited by Deveno; October 15th 2012 at 11:51 AM.
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  4. #4
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    Re: roots with multiplicity > 1 polynomials

    the graph helped with a visual perspective of the problem.

    Thanks, max Jasper
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  5. #5
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    Re: roots with multiplicity > 1 polynomials

    Thanks for your help DEVENO
    i presume the logic to be this.
    since r is a root, r^4 + r + 6 = 0, and 4r^3 + 1 = 0 so, multiplying any constant/variable other than 4 and r will also return a zero. but choosing 'r' and '4' helps us because, we are able to get rid of r^4 helping in a simpler form of a polynomial.

    i was able to follow after that...
    That was of great help.

    Thanks again,
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