solv for x, 6x^3 -7x^2 -12x +7 = 0
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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: The quadratic isn't factorable, so you'll have to use the quadratic formula to get the other two roots.
Isn't it? .
Touché. (Of course I meant factorable over the rationals. )
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f\:6} = \dfrac{\pm 1,\,\pm 7}{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}" alt="\dfrac{Factors\
(Of course I meant factorable over the rationals. 
)
