|
-
Sep 6th, 2002, 11:37 PM
#13
Hyperactive Member
What about:
x^6 + 3x^5 - 41x^4 - 87x^3 + 400x^2 + 444x - 720 = 0
x has 6 real roots, 1, -2, 3, -4, 5, and -6, or unless I'm completely misunderstood, but on another note, if 5 is the limit, and you have x^8, 5 MUST be real, but the other 3 can't be complex because there is always an even number of complex roots.
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|