Question

Answer 1

\(3n^3+10n^2-8=\left(3n+1\right)\left(n^2+3n-1\right)-7\)

Để \(3n^3+10n^2-8⋮3n+1\) thì 

\(7⋮3n+1\)

\(\Rightarrow3n+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

Vì \(n\in Z\)

\(\Rightarrow n\in\left\{0;2\right\}\)

Answer 2

Ta có: \(\dfrac{3n^3+10n^2-8}{3n+1}\)

\(=\dfrac{3n^3+3n^2+7n^2+7n-7n-7-1}{3n+1}\)

\(=3n^2+7n-7+\dfrac{-1}{3n+1}\)

Để đây là phép chia hết thì \(-1⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1\right\}\)

\(\Leftrightarrow3n\in\left\{0;-2\right\}\)

hay n=0