How do you find the zeros of a function $(x^3 - 5x^2+ 4x )/(-4x^2 + 36)$ ?

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Answer 1 · 2015-05-12T00:41:43

First, factor the function:

$f(x) = [x(x^2 - 5x + 4)]/[4(9 - x^2)] = [x(x - 1)(x - 4)]/[4(3 - x)(3 + x)]$

Answers: x = 0; x = 1; and x = 4.

How do you find the zeros of a function (x^3 - 5x^2+ 4x )/(-4x^2 + 36)(x^3 - 5x^2+ 4x )/(-4x^2 + 36)?