for what values of xx^2 - 3x + 2)/(x^2 - 4x + 4) times (x^2 + x – 6)/(x^2 – 1) undefined?

Answer 1

whenever the denominator is 0

+2 for the first one and +1 for the second

Answer 2

An expression is undefined whilst it might in any different case would desire to divide by making use of 0. [ (x^2 - 3x + 2)/(x^2 - 4x + 4) ] * [ (x^2 + x - 6)/(x^2 - a million) ] [ (x^2 - 3x + 2) / (x-2)(x-2) ] * [ (x^2 + x - 6) / (x+a million)(x-a million) ] so which you get branch by making use of 0 whilst x = 2, -a million, or a million in case you tried simplifying this further, you get [ (x-2)(x-a million) / (x-2)(x-2) ] * [ (x-2)(x+3) / (x+a million)(x-a million) ] [ (x-a million) / (x-2) ] * [ (x-2)(x+3) / (x+a million)(x-a million) ] [ (x-a million) ] * [ (x+3) / (x+a million)(x-a million) ] (x+3) / (x+a million) So if we are allowed to simplify the expression, it relatively is in ordinary terms undefined for x= -a million. we are actually not continuously allowed to try this nevertheless; it relies upon on the utility and how the unique function is defined.

Answer 3

(x^2 - 4x + 4) = (X-2)^2

(x^2 – 1) = (x-1)(x+1)

Your product is undefined at the three points, -1, 1, 2, where the denominators are 0.

Note: (x^2 - 3x + 2) = (x-2)(x-1)
and (x^2 + x – 6) = (x+3)(x-2)

but you can't cancel the (x-2) and (x-1) term when you comute the domain. You are only allowed to cancel terms when the denominator is non-zero.

Answer 4

factor both denamenators .. so the first one will be (x-2)^2 and the second one will be (x-1)(x+1) .. so the zeros are 2,1 and -1.. the denomenator cant be 0 because u can't divide anything by zero so the answers are 2,1 and -1. Hope that helps