Question 594307: simplify fully
n!-2(n-2)! / (n-2)(n-2)!
Answer by mamiya(56)   (Show Source):
You can put this solution on YOUR website! n!-2(n-2)! / (n-2)(n-2)!
n!-2(n-2)!/ (n-2)(n-2)! = (n(n-1)(n-2)! -2(n-2)! ) / ( (n-2)(n-2)!)
= (n(n-1) -2 )/ (n-2)
= (n^2 -n -2)/ (n-2)
At this point, we have to look for the factors of n^2-n-2, if it exists
Using the quadratic formula, we get
n^2-n-2=0 , --> n= (1- sqrt( 1-4(-2)) )/2 or n = (1+sqrt( 1-4(-2))) /2
so n= -1 or n=2
this means the factorization of n^2-n-2 is (n-2)(n+1)
so, n!-2(n-2)!/(n-2)(n-2)! = (n+1)(n-2)/ (n-2)
= n+1
so the complete simplification of n!-2(n-2)!/(n-2)(n-2)! is n+1
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