Question

How do you solve `n! = 56(n-2)!`?

Answer 1

I found:
`n_1=8`
`n_2=-7`

Explanation:

We can try this:
`n! =56(n-2)!`
`(n!)/((n-2)!) =56`
then use the property of the factorials and write:
`(n(n-1)(n-2)!)/((n-2)!)=56`
simplify:
`(n(n-1)cancel((n-2)!))/(cancel((n-2)!))=56`
to get:
`n(n-1)=56`
`n^2-n-56=0`
using the Quadratic Formula:
`n_(1,2)=(1+-sqrt(1+224))/2=(1+-15)/2`
so:
`n_1=8`
`n_2=-7`