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Math Help - Polynomials

  1. #1
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    Exclamation Polynomials

    Why is it that the most general form of the remainder in a polynomial is given by R(x) = ax +b?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by xwrathbringerx View Post
    Why is it that the most general form of the remainder in a polynomial is given by R(x) = ax +b?
    Please rephrase this question, at the moment it makes no sense

    RonL
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  3. #3
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    Sori

    The question was : "Explain why the most general form of R(x) i.e. the remainder in the polynomial P(x) = (x+1)(x-4) is given by R(x) = ax+b."
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  4. #4
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    Quote Originally Posted by xwrathbringerx View Post
    Sori

    The question was : "Explain why the most general form of R(x) i.e. the remainder in the polynomial P(x) = (x+1)(x-4) is given by R(x) = ax+b."
    No there is still not enough context, we have established that you are talking about second degree polynomials, but remainder for what?

    RonL
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  5. #5
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    Oops really sori it seems I've been reading the question wrong.

    Explain why the most general form of R(x) is given by R(x) = ax+b.

    WHEN the polynomial P(x) is divided by (x+1)(x-4) where the quotient is Q(x) and the remainder is R(x).

    Sori once more
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  6. #6
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    Quote Originally Posted by xwrathbringerx View Post
    Oops really sori it seems I've been reading the question wrong.

    Explain why the most general form of R(x) is given by R(x) = ax+b.

    WHEN the polynomial P(x) is divided by (x+1)(x-4) where the quotient is Q(x) and the remainder is R(x).

    Sori once more
    If the remainder were a polynomial of degree greater n than 1 then it can be rewriten in the form:

    R(x)=(x+1)(x-4)P_{n-2} (x) +Q_{n-1}(x)

    Where P_{n-2} and Q_{n-1} are polynomials of degree n-2 and no more than n-1 respectivly, so it would not have been the remainder.
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