# solve for factorial.(n+2)!/(n-1)!=336

### 2 Answers | Add Yours

The solutions to n^3+3n^2+2n-336=0 has 6 as a root. so

n=6 is a solution

6*7*8=336

You need to solve for n the factorial equation:

`((n+2)!)/((n-1)!)=336`

You should know that `(n+2)! = (n-1)!*n*(n+1)*(n+2)`

`((n-1)!*n*(n+1)*(n+2))/((n-1)!)=336`

Reducing by (n-1)! yields:

`n*(n+1)*(n+2) = 336`

Opening the brackets yields:

`n^3 + 3n^2 + 2n - 336 = 0`