hello i am working on this study guide for exam 1
determine all the elements in each of the following sets.
{n|0=(n-2)(3n+1)(n-4), n $\displaystyle E$ N}
So all natural number solutions to $\displaystyle (n-2)(3n+1)(n-4) = 0$ or n = 2 and n = 4, for $\displaystyle 3n+1 = 0 $ means n = $\displaystyle \frac{-1}{3} $ which is not a natural number.
so you have $\displaystyle \{2,4\} $