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How do you find the poles of #x/(x^2-3x+2)#?

Answer 1

Benjamin Achtenberg

Poles at #x = 1,2#

#x/(x^2-3x+2) = x/((x-1)(x-2)) #

So poles at #x = 1,2#

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Answer 2

Greyson Brown

To find the poles of the function ( \frac{x}{x^2 - 3x + 2} ), you need to identify the values of ( x ) where the denominator becomes zero, since division by zero is undefined.

The denominator ( x^2 - 3x + 2 ) factors into ( (x - 1)(x - 2) ). Setting this expression equal to zero and solving for ( x ), you get:

[ (x - 1)(x - 2) = 0 ]

This yields ( x = 1 ) and ( x = 2 ) as the values where the denominator becomes zero. Therefore, the poles of the function are ( x = 1 ) and ( x = 2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.