Question 148170
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 1 (the first coefficient):


*[Tex \LARGE q=\pm1]


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



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







Now simplify


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


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