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Math Help - Remainder Theorem

  1. #1
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    Unhappy Remainder Theorem


    Given that f(x) = x^2+px+q , determine the values of the constants p and q so that both
    a) f(x) has a turning point when x=-3 and
    b) the remainder when f(x) is divided by x+2 is 2.
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  2. #2
    MHF Contributor
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    Given that f(x) = x^2+px+q , determine the values of the constants p and q so that both
    a) f(x) has a turning point when x=-3 and
    b) the remainder when f(x) is divided by x+2 is 2.
    " a) f(x) has a turning point when x=-3 "
    Turning point?
    I assume that is a maximum or minimum point on the curve of f(x), so we equate f'(x) to zero:
    f(x) = x^2 +px +q -----------(i)
    f'(x) = 2x +p
    Set f'(x) to zero, and x to -3,
    0 = 2(-3) +p
    p = 6 ---------***
    So, for now,
    f(x) = x^2 +6x +q --------(ii)

    " b) the remainder when f(x) is divided by x+2 is 2. "
    I cannot show the long division here. But (x^2 +6x +q) / (x+2) is
    x+4, remainder (q-8).
    So,
    q-8 = 2
    q = 2+8 = 10 ------***

    Therefore, f(x) = x^2 +6x +10 -----------answer.
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  3. #3
    Senior Member
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    Quote Originally Posted by ticbol
    " b) the remainder when f(x) is divided by x+2 is 2. "
    I cannot show the long division here.
    No need to do the division. The remainder when polynomial f(x) is divided by (x-a) is just the value f(a). So the remainder when x^2 + 6x + q is divided by x+2 is (-2)^2 + 6(-2) + q = q-8.
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