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Math Help - GCD of f(x) and f'(x) (simple question)

  1. #1
    Member elninio's Avatar
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    GCD of f(x) and f'(x) (simple question)

    This seemed simple, but I keep getting ridiculous fractions.

    For example, find the GCD of f(x) and f'(x) over Q of:
    f(x)=x^3 -3x -2

    The quotient becomes 1/3x with a remainder of -2x-2, which can be simplified to x+1.
    From my understanding, x+1 is the GCD. This doesnt seem correct, taking the quotient into consideration. Can anyone check this?
    As a sample, my book has GCD(f(x),f'(x))= x-1 when f(x)=x^4-x^3-x+1. I could not work this question out!


    Also, as a very similar question, if I get a remainder of -2x^2+x, can I simplify this to -2x+1, or does such a simplification not work for GCD's?
    Last edited by elninio; November 25th 2009 at 02:02 PM.
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  2. #2
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    Quote Originally Posted by elninio View Post
    This seemed simple, but I keep getting ridiculous fractions.

    For example, find the GCD of f(x) and f'(x) over Q of:
    f(x)=x^3 -3x -2

    The quotient becomes 1/3x with a remainder of -2x-2, which can be simplified to x+1.
    From my understanding, x+1 is the GCD. This doesnt seem correct,

    But it is: just check that both f(x)\,,\,f'(x) have -1 as one of their roots.


    taking the quotient into consideration. Can anyone check this?

    Also, as a very similar question, if I get a remainder of -2x^2+x, can I simplify this to -2x+1, or does such a simplification not work for GCD's?

    What do you mean "simplify it"? You're dividing it by a non-trivial divisor! That's not simplifying but changing.

    Perhpas what you mean is that, for instance, 2x+2\,,\,x+1 are associate or equivalent elements since one equals the other times a unit.
    In your case, if -2x^2+x=-x(2x-1)=\frac{1}{2}x\!\!\left(x-\frac{1}{2}\right) is the gcd, then you can "factor out" that unit and say the gcd is x\left(x-\frac{1}{2}\right)

    Tonio
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  3. #3
    Member elninio's Avatar
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    Thanks Tonio. That cleared everything up. You're the man!
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  4. #4
    Member elninio's Avatar
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    I'm not sure if I need to make a new thread, but could you check one final question:

    GCD (x^3-2x^2+3x+1,x^3+2x+1) over Z_5.

    I worked it out and got a quotient of 1 with remainder 2x^2+x (=x(2x+1))
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    Quote Originally Posted by elninio View Post
    I'm not sure if I need to make a new thread, but could you check one final question:

    GCD (x^3-2x^2+3x+1,x^3+2x+1) over Z_5.

    I worked it out and got a quotient of 1 with remainder 2x^2+x (=x(2x+1))

    Well, the quotient is correct but the remainder I got is 3x^2+x...and now what? You haven't yet ginished calculating the GCD of those two pol's!
    I advice you to re-read Euclides Method to determine the gcd of two elements in an Euclidean Domain.

    Tonio
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  6. #6
    Member elninio's Avatar
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    Ahhh, I see. Then I would divide the lower degree polynomial by the remainder until I find a remainder of zero then, correct?
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    Quote Originally Posted by elninio View Post
    Ahhh, I see. Then I would divide the lower degree polynomial by the remainder until I find a remainder of zero then, correct?

    Yep, so...go on!

    Tonio
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  8. #8
    Member elninio's Avatar
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    (Disregard this...>SOLVED)
    Last edited by elninio; November 25th 2009 at 04:05 PM.
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