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.