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Math Help - Finding the value of unknown coefficients using a known factor

  1. #1
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    Finding the value of unknown coefficients using a known factor

    This one is really wrecking my head. I've been working on this for hours and I just can't seem to get the correct answer. I'm starting to think there is an error in the textbook, but I've been studying for hours and I think I may just be losing my focus.

    The question is:

    Find the value of p and the value of q if px^3 + qx^2 - 58x -15 is divisible by x^2 + 2x -15.

    So, I've broken the factor down into roots, namely (x + 5) and (x - 3). Using the factor theorem, if f(x) = px^3 + qx^2 - 58x -15, then f(-5) and f(3) should both be equal to 0, right?

    However I've tried and tried and I just can't get the correct values for p an q, which are given as 4 and 9, respectively.

    p(-5)^3 + q(-5)^2 - 58(-5) -15 = 0

    p(-125) + q(25) + 290 -15 = 0

    -125p + 25q + 275 = 0

    25q + 275 = 125p

    q + 11 = 5p

    q = 5p - 11

    ************************
    p(3)^3 + q(3)^2 - 58(3) -15 = 0

    27p + 9q - 174 - 15 = 0

    27p + 9q - 189 = 0

    27p = 189 - 9q

    3p = 21 - q

    p = 7 - q/3

    now, express q in terms of p

    p = 7 - (5p - 11)/3

    3p = 21 - 5p -11

    8p = 10 or p = 10/8 or 1.25. But p should equal 4???

    I have done this over and over and over and over and I just can't get the correct values for p and q. Can someone please tell me where I'm going wrong?

    Thank you. You will be saving me from total brain meltdown.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Finding the value of unknown coefficients using a known factor

    Quote Originally Posted by indymogul View Post
    This one is really wrecking my head. I've been working on this for hours and I just can't seem to get the correct answer. I'm starting to think there is an error in the textbook, but I've been studying for hours and I think I may just be losing my focus.

    The question is:

    Find the value of p and the value of q if px^3 + qx^2 - 58x -15 is divisible by x^2 + 2x -15.

    So, I've broken the factor down into roots, namely (x + 5) and (x - 3). Using the factor theorem, if f(x) = px^3 + qx^2 - 58x -15, then f(-5) and f(3) should both be equal to 0, right?

    However I've tried and tried and I just can't get the correct values for p an q, which are given as 4 and 9, respectively.

    p(-5)^3 + q(-5)^2 - 58(-5) -15 = 0

    p(-125) + q(25) + 290 -15 = 0

    -125p + 25q + 275 = 0

    25q + 275 = 125p

    q + 11 = 5p

    q = 5p - 11

    ************************
    p(3)^3 + q(3)^2 - 58(3) -15 = 0

    27p + 9q - 174 - 15 = 0

    27p + 9q - 189 = 0

    27p = 189 - 9q

    3p = 21 - q

    p = 7 - q/3

    now, express q in terms of p

    p = 7 - (5p - 11)/3

    3p = 21 - 5p -11

    8p = 10 or p = 10/8 or 1.25. But p should equal 4???

    I have done this over and over and over and over and I just can't get the correct values for p and q. Can someone please tell me where I'm going wrong?

    Thank you. You will be saving me from total brain meltdown.
    I can't immediately see the problem with your work, but have you considered simply dividing the two? It's a bit messy but I was able to get the solution with little trouble. (p = 4 and q = 9).

    -Dan
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  3. #3
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    Re: Finding the value of unknown coefficients using a known factor

    Cheers Dan. Can you explain what you mean in a bit more detail? Dividing the two what?

    Thanks.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Finding the value of unknown coefficients using a known factor

    Quote Originally Posted by indymogul View Post
    Cheers Dan. Can you explain what you mean in a bit more detail? Dividing the two what?

    Thanks.
    Good old fashioned long division. (px^3 + qx^2 - 58x - 15)/(x^2 + 2x - 15). You'll have to keep track of the "p's and q's" in order to work it down. Your remainder line (your last line) will look like [-2(q - 2p) + 15p- 58]x + 15(q - 2p - 1). In order to make the two polynomials divide with no remainder you set both coefficients to zero. Two equations, two unknowns.

    -Dan
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