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Division Algorithm for Polynomials in R[x] confusing me

  1. Nov 24, 2014 #1
    I will be using /= to mean 'does not equal'.

    From my textbook:
    Division Algorithm: Let R be any ring and let f(x) and g(x) be polynomials in R[x]. Assume that f(x) /= 0 and that the leading coefficient of f(x) is a unit in R. then unique determined polynomials q(x) and r(x) exist such that
    1) g(x) = q(x)f(x) + r(x)
    2) Either r(x) = 0 or deg[r(x)] < deg[f(x)]

    Can somebody explain to me why the following values for the given functions is invalid?
    q(x) = 2x+3
    f(x) = 3x+4
    r(x) = 3x^2
    g(x) = 9x^2 +17x+12

    These polynomials work and deg[r(x)] = deg[f(x)], more importantly, deg[r(x)] is not less than deg[f(x)].

    I'm assuming part of the reason why what I said is invalid is because f(x) and g(x) aren't polynomails in R[x] or something? Need some math guru to help this noob :D
  2. jcsd
  3. Nov 24, 2014 #2


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    Let's unscramble your post a bit.

    You are given two polynomials
    f(x) = 3x+4
    g(x) = 9x2 +17x+12

    and asked something about two other polynomials which supposedly result from doing the operation g(x) / f(x), where

    q(x) = 2x+3
    r(x) = 3x2

    IOW, q(x) is supposed to represent the quotient of g(x) / (f(x), and r(x) is supposed to represent any remainder from this operation.

    According to 1) above, the first order of business is to check to see if

    1) g(x) = q(x)f(x) + r(x)

    and second, to determine if

    2) Either r(x) = 0 or deg[r(x)] < deg[f(x)]

    Now, q(x) and r(x) as given may result g(x) when inserted into 1) and all of the algebra is carried out. However, the conditions in 2) are not met, since r(x) ≠ 0 and the degree of r(x), namely 2, is not less than the degree of f(x), which is 1. Your statement that deg [r(x)] = deg [f(x)] is incorrect.

    Since conditions 1) and 2) are not satisfied jointly, q(x) and r(x) are not the proper quotient and remainder polynomials which should be produced by the operation g(x) / f(x).
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