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Math Help - remainder theorem

  1. #1
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    remainder theorem

    what do i do for this

    when f(x)= x^3+ax^2+bx-15

    is divided by (x-1)
    when divided by (x-2) there is remainder 35

    find expressions for f(1) and f(2)

    THEN

    set up two simutltaneoues equations and solve to find a and b

    then factorise f(x) completely

    what i do???????????????
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: remainder theorem

    when f(x) is divided by (x-1), then what?
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  3. #3
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    Re: remainder theorem

    when f(x) is divided by (x-1), then what? there is no remainder
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  4. #4
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    Re: remainder theorem

    Quote Originally Posted by Siron View Post
    when f(x) is divided by (x-1), then what?
    there is no remainder
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  5. #5
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    e^(i*pi)'s Avatar
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    Re: remainder theorem

    Quote Originally Posted by kitobeirens View Post
    there is no remainder
    If there is no remainder when f(x) is divided by (x-1) then you know that (x-1) is a factor of f(x) and that f(1)=0

    The remainder theorem says that f(2) = 35 (see spoiler for more info on the remainder theorem)

    Spoiler:
    In more detail the remainder theorem says that f(x) = q(x)g(x) + r(x) where r(x) is of a lower degree than g(x). g(x) itself is the divisor: (x-2) in your case, q(x) the quotient and r(x) the remainder. Subbing this into what you're told: f(x) = q(x)(x-2) + r(x) and since r(x) is of lower degree than g(x) then r(x) is of degree 0 which is just a constant: r(x) = r.

    At the point 2 (where x-2 =0): f(2) = r which in your case is 35
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  6. #6
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    Re: remainder theorem

    Quote Originally Posted by e^(i*pi) View Post
    If there is no remainder when f(x) is divided by (x-1) then you know that (x-1) is a factor of f(x) and that f(1)=0

    The remainder theorem says that f(2) = 35 (see spoiler for more info on the remainder theorem)

    Spoiler:
    In more detail the remainder theorem says that f(x) = q(x)g(x) + r(x) where r(x) is of a lower degree than g(x). g(x) itself is the divisor: (x-2) in your case, q(x) the quotient and r(x) the remainder. Subbing this into what you're told: f(x) = q(x)(x-2) + r(x) and since r(x) is of lower degree than g(x) then r(x) is of degree 0 which is just a constant: r(x) = r.

    At the point 2 (where x-2 =0): f(2) = r which in your case is 35

    so what is a and b ??
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  7. #7
    Super Member Quacky's Avatar
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    Re: remainder theorem

    No, that isn't the way it works. e^{i\pi} has given you the theory in order to enable you to deduce a and b.

    f(2)=35 and
    f(1)=0

    Now it's your turn to do something.
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