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Math Help - find values

  1. #1
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    find values

    hi, i've got the question:

    the polynomials f(x) and g(x) are defined by f(x) = x^3 + px^2 - x + 5, g(x) = x^3 - x^2 + px + 1 where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R

    i got it down to
    R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
    R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.

    can someone show me the way to do it please? thanks
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mark View Post
    hi, i've got the question:

    the polynomials f(x) and g(x) are defined by f(x) = x^3 + px^2 - x + 5, g(x) = x^3 - x^2 + px + 1 where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R

    i got it down to
    R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
    R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.

    can someone show me the way to do it please? thanks
    I would start by dividing each by x-2, then set the remainders equal.
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  3. #3
    Eater of Worlds
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    \frac{x^{3}+px^{2}-x+5}{x-2}=\underbrace{\frac{4p+11}{x-2}}_{\text{remainder}}

    \frac{x^{3}-x^{2}+px+1}{x-2}=\underbrace{\frac{2p+5}{x-2}}_{\text{remainder}}

    4p+11=2p+5\Rightarrow p=-3

    R=-1

    Therefore, the remainder in each case would be \frac{-1}{x-2}

    \frac{x^{3}-x^{2}-3x+1}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}+x-1

    \frac{x^{3}-3x^{2}-x+5}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}-x-3
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  4. #4
    Senior Member pacman's Avatar
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    Galactus, your latexing method is as good as that of soroban, excellent!
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  5. #5
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    ok i understand it now up to \frac {-1}{x - 2}, when you get to that, wouldn't x have to be 3 for the final answer to be -1? how do you know the value of x?
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  6. #6
    Eater of Worlds
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    Yes, p=-3. Sub that in and we get R=-1
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