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

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

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Answer 1 · 2015-05-12T00:41:43

First, factor the function:

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

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

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

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How do you find the zeros of a function x3−5x2+4x−4x2+36(x^3 - 5x^2+ 4x )/(-4x^2 + 36)?

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