Question 131238
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 12 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm3, \pm4, \pm6, \pm12]


Now let's list the factors of 2 (the first coefficient):


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


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{2}{1}, \frac{2}{2}, \frac{3}{1}, \frac{3}{2}, \frac{4}{1}, \frac{4}{2}, \frac{6}{1}, \frac{6}{2}, \frac{12}{1}, \frac{12}{2}, \frac{-1}{1}, \frac{-1}{2}, \frac{-2}{1}, \frac{-2}{2}, \frac{-3}{1}, \frac{-3}{2}, \frac{-4}{1}, \frac{-4}{2}, \frac{-6}{1}, \frac{-6}{2}, \frac{-12}{1}, \frac{-12}{2}]







Now simplify


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


*[Tex \LARGE  1, \frac{1}{2}, 2, 3, \frac{3}{2}, 4, 6, 12, -1, \frac{-1}{2}, -2, -3, \frac{-3}{2}, -4, -6, -12]