Question 187061
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 15 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm3, \pm5, \pm15]


Now let's list the factors of 3 (the first coefficient):


*[Tex \LARGE q=\pm1, \pm3]


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



*[Tex \LARGE \frac{1}{1}, \frac{1}{3}, \frac{3}{1}, \frac{3}{3}, \frac{5}{1}, \frac{5}{3}, \frac{15}{1}, \frac{15}{3}, \frac{-1}{1}, \frac{-1}{3}, \frac{-3}{1}, \frac{-3}{3}, \frac{-5}{1}, \frac{-5}{3}, \frac{-15}{1}, \frac{-15}{3}]






Now simplify


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


*[Tex \LARGE  1, \frac{1}{3}, 3, 5, \frac{5}{3}, 15, -1, -\frac{1}{3}, -3, -5, -\frac{5}{3}, -15]