Question

Determine all the elements in each of the following sets

{n|0=(n-2)(3n+1)(n-4), n (is an element of) N}

Answer 1

\[\left\{ n | 0=(n-2)(3n+1)(n-4), n \in N\right\}\]

Answer 2

oh um hi i kinda got the answer which n = 2 and n = 4
but where does the 3n+1 fit in?

Answer 3

i guess it doesn't since if \(3n+1=0\iff n=-\frac{1}{3}\) which is not an integer

Answer 4

how did you even get - \[\frac{ 1 }{ 3 }\]

Answer 5

i solved \(3n+1=0\) for \(n\)

Answer 6

how does that turn in to a fraction?