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Math Help - rational roots

  1. #1
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    rational roots

    f(x)=3x^3-26x^2+52x-24

    write down all the solutions of the equation f(x)=0
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  2. #2
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    Quote Originally Posted by outlaw View Post
    f(x)=3x^3-26x^2+52x-24

    write down all the solutions of the equation f(x)=0
    If there exist integer solution they must be factors of 24. Determine all factors of
    24 : 1, 2, 3, 4, 6, 8, 12, 24

    Check if one of these 16 fctors is a solution of the equation. I've found x = 2.

    Use synthetic division to cancel out the factor (x - 2). I've got:

    (3x^3-26x^2+52x-24) \div (x-2)=3x^2-20x+12

    Solve the quadratic

    3x^2-20x+12 = 0 for x to get the last 2 solutions of the equation
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  3. #3
    Member Jonboy's Avatar
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    Hey outlaw!

    Solve: f (x) = 3x^3 - 26x^2 + 52x - 24 for  f (0).
    Looks like you can factor out a 3, but 3 doesn't go into 52 evenly.
    And it looks like you can't Factor by grouping.

    So use the Rational Roots Theorem or Rational Zeros Theorem.

    With that, you'll have: \frac{p}{q} = \frac{\pm 1\,,\,\pm2\,,\,\pm 4 \,,\,\pm 6\,,\,\pm8\,,\,\pm12\,,\,\pm24}{\pm 1\,,\,\pm 3}

    So plug all possible values \frac{p}{q} into the equation to see which ones give you zero.

    When you do that, I get f(2), f(6) and f(\frac{2}{3}) give you zero.
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