Question 475525
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You are working way too hard.  You properly derived the quadratic equation and obtained the correct solutions.  You discard the negative root because the factorial function is not defined for negative integers.  That leaves you with 5 as the only answer to the problem.


No, the answer cannot be 0 because (0 - 2)! for the same reason that you excluded the -4, i.e. the factorial function is not defined for negative integers.  The LHS of your equation does NOT go to zero, first because of the undefined part of the denominator and second because your numerator of 0! = 1 NOT 0.  See the definition of Factorial:


<a href="http://en.wikipedia.org/wiki/Factorial">Wikipedia Factorial</a>


No, the answer cannot be n = 2 because that would give you a zero factor in the denominator.


Take the term "cancelling out" and remove it from your mathematical vocabulary.  You are dividing identical terms so as to achieve an overall multiplier of the fraction of 1.  What your book is trying to tell you is that you cannot have a zero factor in your denominator.  If n = 2, then n - 2 = 0.


Hope that helps.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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