Question

Solve the equation : $3n^3+4n^2+n=0$

Answer 1

$3n^3+4n^2+n=0$

$\Rightarrow n[3n^2+4n+1]=0$

$\Rightarrow 3n^2+4n+1=0$ n=0

$\Rightarrow 3n^2+3n+n+1=0$

$\Rightarrow 3n[n+1]+1[n+1]=0$

$\Rightarrow (n+1)(3n+1)=0$

$\Rightarrow$ n=-1 or 3n=-1

$\Rightarrow$ n=-1 or $n=-\frac{1}{3}$

$\therefore n=0,-1,-\frac{1}{3}$