Question 113801

Any rational zero can be found through this formula


*[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 of {{{12x^4-67x^3+108x^2-47x+6}}}):


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


Now let's list the factors of 12 (the first coefficient of {{{12x^4-67x^3+108x^2-47x+6}}}):


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


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







Now simplify


These are all the distinct rational zeros of the function that could occur (remember these are all <b>possible</b> zeros)


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