Question 145526
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 10 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm5, \pm10]


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{5}{1}, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \frac{10}{1}, \frac{10}{2}, \frac{10}{4}, \frac{10}{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{-5}{1}, \frac{-5}{2}, \frac{-5}{4}, \frac{-5}{8}, \frac{-10}{1}, \frac{-10}{2}, \frac{-10}{4}, \frac{-10}{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, 5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, 10, -1, \frac{-1}{2}, \frac{-1}{4}, \frac{-1}{8}, -2, -5, \frac{-5}{2}, \frac{-5}{4}, \frac{-5}{8}, -10]