My solution below proves all possible factorings where one factor is
linear. It doesn't rule out the possibility of higher degree polynomial factors.
The Rational Zero Theorem says any rational (in this case integer) zeros may only be at 1, 1, 2, 2, 3, 3, 6, or 6.
Synthetic division on the first three terms 1 3 2 by each of these zeros give quotient and remainder as follows:
1: 1 2 0 0 0 0...
1: 1 4 6 6 6 6...
2: 1 1 0 0 0 0...
2: 1 5 12 24 48 96...
3: 1 0 2 6 18 54...
3: 1 6 20 60 180 540...
6: 1 3 20 120 720...
6: 1 9 56 336...
What we need to look for is a 6. If a 6 is inserted above it the sum would be 0 and we'd have our factor.
This happens only at every other position for the division with 1.
So (x+1) is a factor for every odd degree.
The answer would seem to be that n can be any odd ≥ 3.
(The method above also shows how replacing 6 with 6 would change the solution to n can be 3 or and even ≥ 4)

Posted by Jer
on 20150323 14:05:30 