What is the zero of x^3+1/x^2-1 (my answer is -1. Is that correct?) please help

Answer 1

If you meant
(x^3+1)/(x^2-1)
then if x = -1, your expression becomes 0/0 which is "indeterminate".
notice that (x^3+1)/(x^2-1)
= (x+1)(x^2 + x + 1)/((x+1)(x-1))
= (x^2 + x + 1)/(x-1)

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%5E2-1%29
will show you.
the expression actually reduces to
y = (x^2 + x + 1)/(x-1) , which has a vertical asymptote at x=1

the original function
y = (x^3 + 1)(/(x^2-1) looks exactly the same except it would have a "hole" or missing point at (-1, -1/2)

Answer 2

Thank u so much!