Posted by **Muffy** on Saturday, October 31, 2009 at 1:57pm.

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?

- Pre Calc -
**bobpursley**, Saturday, October 31, 2009 at 2:06pm
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.

- Pre Calc -
**Muffy**, Saturday, October 31, 2009 at 2:17pm
So I was correct the answer is i right?

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