Would this be x^6+ax^5+x^4-x^3+3x-a2
Assuming this is a binomial expansion ie all terms are a, then
(-a)^6=-2a or a=-(2)^1/5.
However, if this is a binomial expansion. then this says 6a^5=3 and that is not true if a =-(2)^1/5
So lets iterate and let a=1, then assume it is 1 of six possible roots.
Then we have (-1)^6 -a(-1)^5+(-1)^4-(-1)^3+3(-1) -2a=1+a+1+1-3-2a=0
-a=0 or -1=0 which means that -1 is not a root and a is therefore not 1.
Lets test a=-1............so we have (1)^6-a(1)^5+(1)^4-(1)^3+3(1)-2a=
1-a+1-1+3-2a=0 and 4+3 does not equal 0
Lets try a=2
We then have(-2)^6-a(-2)^5+(-2)^4-(-2)^3+3(-2)-2a=64+32a+16-8-6-2a which doesn't equal 0..
Without going into specifics, letting a equal any integer larger than 2 will not cause a 0.
I don't believe there is an integer solution for this. I used to have a calculator that would solve this. This is an iterative problem and is solved by converging on the answer.
I don't have the time but would guess a is between 0 and 1. Just an educated guess. :D
Maybe I don't understand what equation you are using.
This cant be a binomial expansion and it converges above 2 and less than -2. This needs to be iterated until you converge on an answer. Unless someone has a graphing calculator :D.