Posted by Meron on .
What is the zero of x^3+1/x^21 (my answer is 1. Is that correct?) please help

Math 
Reiny,
If you meant
(x^3+1)/(x^21)
then if x = 1, your expression becomes 0/0 which is "indeterminate".
notice that (x^3+1)/(x^21)
= (x+1)(x^2 + x + 1)/((x+1)(x1))
= (x^2 + x + 1)/(x1)
I don't know if you know anything about limits, but
the limit of the above , as x >  1
= (1  1 + 1)/(2) = 1/2
There are actually no zeros for this graph,
as
http://www.wolframalpha.com/input/?i=%28x%5E3%2B1%29%2F%28x%5E21%29
will show you.
the expression actually reduces to
y = (x^2 + x + 1)/(x1) , which has a vertical asymptote at x=1
the original function
y = (x^3 + 1)(/(x^21) looks exactly the same except it would have a "hole" or missing point at (1, 1/2) 
Math 
Meron,
Thank u so much!