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Math Help - Remainder theorum

  1. #1
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    Remainder theorum

    Determine the remainder when x-6x+x-5 is divided by:
    a. X+2
    b. X-3


    Hi I was wondering if someone can help me with these revision questions.
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    Re: Remainder theorum

    Quote Originally Posted by Andrew187 View Post
    Determine the remainder when x-6x+x-5 is divided by:
    a. X+2
    b. X-3.

    If f(x)=x^3-6x^2+x-5 then if f(a)=0 then (x-a) is a factor, so it divides f(x).

    Thus what is f(-2)~\&~f(3)~?
    Last edited by Plato; April 15th 2013 at 03:03 AM.
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    Re: Remainder theorum

    f(x) = x - 6x + x - 5

    (x+2)= f(-2) = (-2) - 6(-2) + (-2) - 5 =

    -8 - 24 - 2 - 5 = -39


    Is this correct? I can't work out the answer for the first formula
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    Re: Remainder theorum

    Quote Originally Posted by Plato View Post
    If f(x)=x^2-6x^2+x-5 then if f(a)=0 then (x-a) is a factor, so it divides f(x).

    Thus what is f(-2)~\&~f(3)~?
    That actually doesn't help the OP.

    The remainder theorem states that for a polynomial function P(x) is divided by (x - a), the remainder is equal to P(a).
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    Re: Remainder theorum

    Quote Originally Posted by Andrew187 View Post
    f(x) = x - 6x + x - 5
    (x+2)= f(-2) = (-2) - 6(-2) + (-2) - 5 =
    -8 - 24 - 2 - 5 = -39
    Is this correct? I can't work out the answer for the first formula
    Yes that is correct.
    Thus (x+2) does not divide f(x). So -39 must be the remainder.

    Surely you can find (3)^3-6(3)^2+(3)-5=~?
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    Re: Remainder theorum

    Quote Originally Posted by Plato View Post
    Yes that is correct.
    Thus (x+2) does not divide f(x). So -39 must be the remainder.

    Surely you can find (3)^3-6(3)^2+(3)-5=~?
    its -29

    But how come your formula is different to my first formula have I wrote it wrong?
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    Re: Remainder theorum

    Quote Originally Posted by Andrew187 View Post
    But how come your formula is different to my first formula ?
    How is it different?
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    Re: Remainder theorum

    Quote Originally Posted by Plato View Post
    Yes that is correct.
    Thus (x+2) does not divide f(x). So -39 must be the remainder.

    Surely you can find (3)^3-6(3)^2+(3)-5=~?
    Is that sum for (x-3)?
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    Re: Remainder theorum

    Quote Originally Posted by Andrew187 View Post
    Is that sum for (x-3)?

    That is the remainder when f(x)=x^3-6x^2+x-5 is divided by (x-3).
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    Re: Remainder theorum

    Quote Originally Posted by Plato View Post
    Yes that is correct.
    Thus (x+2) does not divide f(x). So -39 must be the remainder.

    Surely you can find (3)^3-6(3)^2+(3)-5=~?
    27-24+3-5=1 is this correct?
    or is it this 27-72+3-5= -43
    Last edited by Andrew187; April 15th 2013 at 04:13 AM.
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