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

  • Math -

    If you meant
    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,
    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)

  • Math -

    Thank u so much!

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