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Thread: Hcf polynomials

  1. #1
    Junior Member
    Aug 2007

    Hcf polynomials

    Hi there,
    let f(X) = 2X^4 + 5X^3 + 8X^2 + 7X + 4 and g(X) = 2X^2 + 3X + 3
    the HCF(f, g) = 1.

    but is there a special way to do this kind of question... or do i have to compute manually by long division?


    now let  \alpha\epsilon C satisfy f(\alpha) = 0.
    Find a,b,c,d  \epsilon Q such that :

    1/g(\alpha) = a + b\alpha + c\alpha^2 + d\alpha^3

    any clue on how to go about with this one please? ill try it out for myself once i have an idea on what route to follow?

    thnx a great deal guys
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    If HCF(f,g)=1 then exists two polynomials h_1,h_2 such as

    In this case, applying Euclid's algorithm, we have:
    f=(X^2+X+1)g+X+1 (1)
    so, HCF(f,g)=1.

    Then, \displaystyle X+1=\frac{g-2}{2X+1}
    Plug in (1) and we have (2X+1)f-(2x^3+3X^2+3X+2)g=-2
    Divide by -2: \left(-X-\frac{1}{2}\right)f+\left(X^3+\frac{3}{2}X^2+\frac  {3}{2}X+1\right)g=1

    Now, for x=\alpha we have \left(\alpha^3+\frac{3}{2}\alpha^2+\frac{3}{2}\alp  ha+1\right)g(\alpha)=1\Leftrightarrow\frac{1}{g(\a  lpha)}=\alpha^3+\frac{3}{2}\alpha^2+\frac{3}{2}\al  pha+1
    Then a=1,b=\frac{3}{2},c=\frac{3}{2},d=1
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