How do you find the zeros of a function $g (x) = \frac{x^{3} - 16 x}{4 x^{2} - 4 x}$ $g(x)=(x^3 -16x)/(4x^2 - 4x)$ ?

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How do you find the zeros of a function $g (x) = \frac{x^{3} - 16 x}{4 x^{2} - 4 x}$ $g(x)=(x^3 -16x)/(4x^2 - 4x)$ ?

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Answer 1 · 2015-05-12T00:45:56

First factor the function:

$g (x) = \frac{x (x^{2} - 16)}{4 x (x - 1)} = \frac{x (x - 4) (x + 4)}{4 x (x - 1]}$ $g(x) = [x(x^2 - 16)]/[4x(x - 1)] = [x(x - 4)(x + 4)]/[4x(x - 1]$

The zeros are: x = 0; x = 4; and x = -4

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How do you find the zeros of a function $g (x) = \frac{x^{3} - 16 x}{4 x^{2} - 4 x}$ $g(x)=(x^3 -16x)/(4x^2 - 4x)$ ?

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How do you find the zeros of a function $g (x) = \frac{x^{3} - 16 x}{4 x^{2} - 4 x}$ $g(x)=(x^3 -16x)/(4x^2 - 4x)$ ?