Question 156870
Possible Zeros:



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 10 (the first coefficient):


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


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}{5}, \frac{1}{10}, \frac{2}{1}, \frac{2}{2}, \frac{2}{5}, \frac{2}{10}, \frac{3}{1}, \frac{3}{2}, \frac{3}{5}, \frac{3}{10}, \frac{6}{1}, \frac{6}{2}, \frac{6}{5}, \frac{6}{10}, -\frac{1}{1}, -\frac{1}{2}, -\frac{1}{5}, -\frac{1}{10}, -\frac{2}{1}, -\frac{2}{2}, -\frac{2}{5}, -\frac{2}{10}, -\frac{3}{1}, -\frac{3}{2}, -\frac{3}{5}, -\frac{3}{10}, -\frac{6}{1}, -\frac{6}{2}, -\frac{6}{5}, -\frac{6}{10}]







Now simplify


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


*[Tex \LARGE  1, \frac{1}{2}, \frac{1}{5}, \frac{1}{10}, 2, \frac{2}{5}, 3, \frac{3}{2}, \frac{3}{5}, \frac{3}{10}, 6, \frac{6}{5}, -1, -\frac{1}{2}, -\frac{1}{5}, -\frac{1}{10}, -2, -\frac{2}{5}, -3, -\frac{3}{2}, -\frac{3}{5}, -\frac{3}{10}, -6, -\frac{6}{5}]