Question 138659
Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let's list the factors of -6 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm3, \pm6]


Now let's list the factors of 8 (the first coefficient):


*[Tex \LARGE q=\pm1, \pm2, \pm4, \pm8]


Now let's divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{2}{1}, \frac{2}{2}, \frac{2}{4}, \frac{2}{8}, \frac{3}{1}, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{6}{1}, \frac{6}{2}, \frac{6}{4}, \frac{6}{8}, \frac{-1}{1}, \frac{-1}{2}, \frac{-1}{4}, \frac{-1}{8}, \frac{-2}{1}, \frac{-2}{2}, \frac{-2}{4}, \frac{-2}{8}, \frac{-3}{1}, \frac{-3}{2}, \frac{-3}{4}, \frac{-3}{8}, \frac{-6}{1}, \frac{-6}{2}, \frac{-6}{4}, \frac{-6}{8}]





Now simplify


These are all the distinct rational zeros of the function that could occur


*[Tex \LARGE  1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, 2, 3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, 6, -1, \frac{-1}{2}, \frac{-1}{4}, \frac{-1}{8}, -2, -3, \frac{-3}{2}, \frac{-3}{4}, \frac{-3}{8}, -6]