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Thread: polynomials [another one]

  1. #1
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    polynomials [another one]

    You have 2 polynomials

    $\displaystyle p(x) = x^4 + 1$
    $\displaystyle q(x) = x^3 - 1$

    $\displaystyle d(x) $ is their greatest common divisor.

    a) Find $\displaystyle d(x)$
    b) Find such $\displaystyle f(x) $ and $\displaystyle g(x)$ so that the following is true:

    $\displaystyle p(x)f(x) + q(x)g(x) = d(x) $


    For a) i tried with Euclid's algorithm but it didn't give me any good results (or maybe i did it wrong,, is it done differently than with numbers?)

    no clue for b).

    can anyone help?
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  2. #2
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    that's what i tried:

    $\displaystyle (x^4 + 1) = a \cdot (x^3 - 1) + r(x)$

    I need to find r(x) ->

    $\displaystyle \frac{(x^4 + 1)}{x^3 - 1} = x + \frac{1 + x}{x^3 - 1}$

    a = x
    $\displaystyle \frac{1 + x}{x^3 - 1} = r(x)$

    I continue with algorthm:

    $\displaystyle x^3 - 1 = b \cdot \frac{1 + x}{x^3 - 1} + r_1(x)$

    I need to find $\displaystyle r_1(x)$ ->

    $\displaystyle \frac{x^3 - 1}{\frac{1 + x}{x^3 - 1}} = \frac{x^6 - 2x^3 + 1}{x +1} = x^5 - x^4 + x^3 - x^2 + x - 1 + \frac{2}{x + 1}$

    $\displaystyle \frac{2}{x + 1} = r_1(x) $
    $\displaystyle x^5 - x^4 + x^3 - x^2 + x - 1 = b$

    Continuing with algorithm:

    $\displaystyle \frac{1 + x}{x^3 - 1} = c \cdot r_1(x) + r_2(x) \Rightarrow $


    $\displaystyle \frac{1 + x}{x^3 - 1} = c \cdot \frac{2}{x + 1} + r_2(x) = $

    I need to find $\displaystyle r_2(x)$

    $\displaystyle \frac{\frac{1 + x}{x^3 - 1}}{\frac{2}{x + 1}} = \frac{x^2 + 2x + 1}{2x^3 - 2} = c + r_3(x)$

    But that's where i get confused by it.. the Power of the first one is smaller than second one -> can't divide..

    What am i doing wrong?
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  3. #3
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    Smile

    here's what i think a hint,
    $\displaystyle x^4+1=x(x^3-1)+(x+1).$
    $\displaystyle x^3-1=(x+1)(x^2-x+1)-2.$
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  4. #4
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    oops
    i feel stupid now. :P

    so the greatest common divisor is $\displaystyle - \frac{x^2 - x +1}{2}$ then?

    any hints for b)? :P
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