SOLUTION: Respected Sir, Please help me to solve this question: If(n+2)!=60[(n-1)!],find n. Thank you in anticipation.

Question 741213: Respected Sir,
Please help me to solve this question:
If(n+2)!=60[(n-1)!],find n.
Thank you in anticipation.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
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

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