How do you find the poles of $\frac{x}{x^{2} - 3 x + 2}$ $x/(x^2-3x+2)$ ?

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Answer 1 · 2017-02-18T00:35:09

Poles at $x = 1, 2$ $x = 1,2$

$\frac{x}{x^{2} - 3 x + 2} = \frac{x}{(x - 1) (x - 2)}$ $x/(x^2-3x+2) = x/((x-1)(x-2))$

So poles at $x = 1, 2$ $x = 1,2$

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