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Math Help - Solving Equation

  1. #1
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    Solving Equation

    solv for x, 6x^3 -7x^2 -12x +7 = 0
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  2. #2
    Senior Member eumyang's Avatar
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    Using the Rational Zeros Test:
    f\:7}{Factors\f\:6} = \dfrac{\pm 1,\,\pm 7}{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}" alt="\dfrac{Factors\f\:7}{Factors\f\:6} = \dfrac{\pm 1,\,\pm 7}{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}" />

    ... and guess-and-check with these potential rational zeros, I see that 1/2 is a zero. So the polynomial factors to:
    6x^3 - 7x^2 - 12x + 7 = (2x - 1)(3x^2 - 2x - 7)

    The quadratic isn't factorable, so you'll have to use the quadratic formula to get the other two roots.
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  3. #3
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    Isn't it?

    (2x - 1)(3x^2 -2x - 7) = 3(2x -1)\left(x^2 - \frac{2}{3}x - \frac{7}{3}\right)

     = 3(2x - 1)[x^2 - \frac{2}{3}x + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 - \frac{7}{3}\right]

     = 3(2x - 1)\left[\left(x-\frac{1}{3}\right)^2-\frac{22}{9}\right]

     = 3(2x-1)\left(x-\frac{1}{3}+\frac{\sqrt{22}}{3}\right)\left(x-\frac{1}{3}-\frac{\sqrt{22}}{3}\right).
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  4. #4
    Senior Member eumyang's Avatar
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    Touché. (Of course I meant factorable over the rationals. )
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