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Math Help - Highest common factor, Polynomial division

  1. #1
    Super Member craig's Avatar
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    Highest common factor, Polynomial division

    Hi, think this is the correct forum, apologies if not.

    Find the highest common factor of the polynomials
    x^5 - 2x^4 - 3x^3 + 13x^2 -16x + 6 and x^4 - 2x^3 - 2x^2 + 8x - 8.

    Hence factorise x^4 - 2x^3 - 2x^2 + 8x - 8 into irreducible polynomials in R[x].

    First I started off by factorising, by long division I got the following:

    x^5 - 2x^4 - 3x^3 + 13x^2 -16x + 6 = x(x^4 - 2x^3 - 2x^2 + 8x - 8) + (-x^3 +5x^2-8x+6)

    Not sure where to go from here. I'm ok with the Euclidean algorithm but not sure how to apply it here?

    Thanks in advance
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  2. #2
    Super Member craig's Avatar
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    Edit just seen that I've not fully completed the algorithm yet! Think I can manage this..
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  3. #3
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    I wondered about that- because what you wrote involved neither factoring nor long division!
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  4. #4
    Super Member craig's Avatar
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    Yeh sorry about that. Posted the question and then realised that I'd only done the first bit... oops.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    It's often also wise before you start to do a rational root test:

    Let f(x)=x^4-2x^3-2x^2+8x-8

    Suppose f(x) =0 has rational roots then it either has to be x=\pm 1, \pm 2,\pm 4,\pm 8.

    I checked for you and notice that f(2)=f(-2) = 0 ! This means you can allready factor out f(x)/(x-2)(x+2) = x^2-2x+2.

    For the other function g we have g(\pm 2)\neq 0. This means that you only have to check whether the roots of x^2-2x+2 are roots of g(x) and you're done.

    Edit: I can better put it like this. It now follows that gcd (g(x),f(x)) = gcd( g(x), x^2-2x+2)
    Last edited by Dinkydoe; January 22nd 2010 at 08:13 AM.
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  6. #6
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    Your first step was the first step in Euclidean factorization:

    <br /> <br />
x^5 - 2x^4 - 3x^3 + 13x^2 -16x + 6 = x(x^4 - 2x^3 - 2x^2 + 8x - 8) + (-x^3 +5x^2-8x+6)<br />

    Now you have to do this again, but with the polynomials:
    <br /> <br />
x^4 - 2x^3 - 2x^2 + 8x - 8 =  (-x)(-x^3 +5x^2-8x+6) + remainder<br />

    Keep going until the remainder is 0.
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  7. #7
    Senior Member Dinkydoe's Avatar
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    Your first step was the first step in Euclidean factorization:



    Now you have to do this again, but with the polynomials:


    Keep going until the remainder is 0.
    That is clearly impossible: Only possible when g(x) = f(x)\cdot p(x) For some linear factor p(x). This is not the case here, there will allways be a remainder.
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