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Math Help - The Remainder Theorem

  1. #1
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    The Remainder Theorem

    The remainder theorem says when a polynomial function is divided by ax-b, the remainder is f(b/a). But if you take a function and divided it by 2x-1, for example, the remainder will be obviously different than when divided by 4x-2, which is 2x-1 multiplied by two. f(2/4) is the same as f(1/2), so the result when subbed into f(x) should be the same, but their remainders are obviously different. Can anybody clarify?
    Thanks in advance!
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  2. #2
    Grand Panjandrum
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    Re: The Remainder Theorem

    Quote Originally Posted by Dragon08 View Post
    The remainder theorem says when a polynomial function is divided by ax-b, the remainder is f(b/a). But if you take a function and divided it by 2x-1, for example, the remainder will be obviously different than when divided by 4x-2, which is 2x-1 multiplied by two. f(2/4) is the same as f(1/2), so the result when subbed into f(x) should be the same, but their remainders are obviously different. Can anybody clarify?
    Thanks in advance!
    P(x)=(4x-2)Q(x)+R

    and

    P(x)=(2x-1)Q'(x)+R'

    Q'(x)=2Q(x)

    and R=R'

    CB
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  3. #3
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    Re: The Remainder Theorem

    How did you know that Q'(x)=2Q(x)?
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  4. #4
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    Re: The Remainder Theorem

    Quote Originally Posted by Dragon08 View Post
    How did you know that Q'(x)=2Q(x)?

    When doing polynomial long division of P(x) by (4x-2) and by (2x-1) the quotients differ by a factor of 2.

    Also:

    P(x)=(4x-2)Q(x)+R=(2x-1) [2Q(x)] +R

    But R is a pure number, so:

    Q'(x)=2Q(x) and R=R'

    CB
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  5. #5
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    Re: The Remainder Theorem

    Oh, okay. Thanks. So this only works for fraction divisors right? For example, 10 divided by 3 and 10 divided by 6 don't get the same remainders.
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