Question 193568
You have to be careful when it comes to using the graphing calculator to find the solutions. It turns out that 2.414679875... is a solution (not 2.4). So {{{12/5}}} is NOT a solution.



a)



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 35 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm5, \pm7, \pm35]


Now let's list the factors of 5 (the first coefficient):


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


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



*[Tex \LARGE \frac{1}{1}, \frac{1}{5}, \frac{5}{1}, \frac{5}{5}, \frac{7}{1}, \frac{7}{5}, \frac{35}{1}, \frac{35}{5}, \frac{-1}{1}, \frac{-1}{5}, \frac{-5}{1}, \frac{-5}{5}, \frac{-7}{1}, \frac{-7}{5}, \frac{-35}{1}, \frac{-35}{5}]







Now simplify


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


*[Tex \LARGE  1, \frac{1}{5}, 5, 7, \frac{7}{5}, 35, -1, -\frac{1}{5}, -5, -7, -\frac{7}{5}, -35]



which can be sorted and rewritten as


*[Tex \LARGE  \pm \frac{1}{5}, \pm 1, \pm \frac{7}{5}, \pm 5, \pm 7, \pm 35]



So you have the correct possible rational roots.



b)


From here, simply perform synthetic division using all of the possible zeros to determine which possible roots are actual roots.