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Thread: Hcf polynomials

  1. #1
    Junior Member
    Aug 2007

    Hcf polynomials

    Hi there,
    let $\displaystyle f(X) = 2X^4 + 5X^3 + 8X^2 + 7X + 4$ and $\displaystyle g(X) = 2X^2 + 3X + 3$
    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 $\displaystyle \alpha\epsilon C$ satisfy $\displaystyle f(\alpha) = 0$.
    Find a,b,c,d $\displaystyle \epsilon Q$ such that :

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

    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 $\displaystyle h_1,h_2$ such as
    $\displaystyle h_1f+h_2g=1$

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

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

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