Solve for n:

n^3+3n^2+2n-60=0

Ans:3

If (n+2)! = 60[(n-1)!], n=?
(n+2)! = 60* ( (n-1)! )
(n+2)! = (n+2) * (n+1) * (n) * (n-1)! 
{ (n+2) * (n+1) * (n) * (n-1)! } = 60* ( (n-1)! )
=> { (n+2) * (n+1) * (n) * (n-1)! } = 60*( (n-1)! ) 
=> n3 + 3n2 +2n - 60 = 0
for n=3 we get 0 Substitute n = 3  =>( n3 + 3n2 +2n - 60 ) / [n-3]
quotient= n2 +6n + 20 
=> (n-3) , (n2 +6n + 20)
n3 + 3n2 +2n - 60 = 0
=> (n-3) * (n2 +6n + 20) = 0 
=> (n-3) = 0 or (n2 +6n + 20) = 0
So n = 3