are you looking for factors of x^793?
or i^793, where i is the sqrt(-1)?
factors of x^793 are very numerous. x^793 is not a prime number.
i^793=i^792 * 1
Now consider
i^2=-1
i^4=1
i^6=-1
so the conclusion is that i^(4n-2) is -1
otherwise, i to even is 1
WEll, is 792=4n-2
4n=794, or n is not an integer, so
i^792=1
finally, i^792*i= i and it has only one factor, i.
My question is I^793
I think the answer is just I because it is the only number that can go into it. Is that correct?
2 answers
So I was correct the answer is i right?
1 answer