Question 1148151
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In your rational function, the numerator is equal to  (x-4)*(x+3),

while the denominator is  (x-2)*(x+3).


Therefore, this rational function is equal to

     {{{(x-4)/(x-2)}}} > 0         (1)

everywhere, where the last rational function is defined, except x= -3, where the original rational function is not defined, at all.



Thus our task is to find the solutions of the inequality (1) everywhere, except x= -3.



The function (1) has two critical points  x= 4 and x= 2, where the numerator and denominator change their signs.



In the interval  x < 2, both the numerator and denominator of the function (1) are negative;  so the function (1) is positive.



In the interval 2 < x < 4, the numerator of (1) is negative, while the denominator is positive, so the function (1) is negative.



In the interval x > 4,  both the numerator and denominator of (1) are positive, so the function (1) is positive.



Now I am ready to present the 



<U>ANSWER</U> :  The original function is positive in the intervals 

                 ({{{-infinity}}},{{{-3}}}), (-3,2), and ({{{4}}},{{{infinity}}}).
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Solved.