well, one thing you could do is multiply correctly:
n(n-1)(n-2) = n^3 - 3n^2 + 2n
so,
n^3 - 3n^2 + 2n = 120
n^3 - 3n^2 + 2n - 120 = 0
That could be hard, but try some integer roots. A little synthetic division yields
n^3 - 3n^2 + 2n - 120 = (n-6)(n^2+3n+20)
P(n,3)=5! Solve for n?? Help, please!
I did n!/(n-3)!=5!
n(n-1)(n-2)(n-3)!/(n-3)!=5!
cancelled out (n-3)!
was left with n(n-1)(n-2)=5!
n^2-n(n-2)=5!
n^3-n^2=5!
n^3-n^2-120
What do i do???
1 answer