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Mathematics 7 Online
OpenStudy (anonymous):

Solve (x-1)^2(x+2)(x-3)/x-2 >0

OpenStudy (anonymous):

Absolute Value and Inequalities

OpenStudy (valpey):

\[\frac{(x-1)^{2}(x+2)(x-3)}{(x-2)} > 0\] correct?

OpenStudy (valpey):

\[\frac{(x-1)^{2}(x+2)(x-3)}{(x-2)} > 0\] If (x-2) is positive then: \[(x-1)^{2}(x+2)(x-3) > 0\] which is only going to be true for x > 3. However, if (x-2) is negative then: \[(x-1)^{2}(x+2)(x-3) < 0\] which is only true when x is between -2 and 2 because \((x-1)^2\) is always positive, \((x+2)\) is positive for x > -2, and \((x-3)\) is negative for all x < 2.

OpenStudy (valpey):

Except for x = 1. There it is also zero.

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