((x+12)(x-2))/(x-1) >= 0

First we find the domain for the inequality.

==> The inequality if not defined when **x = 1**.

==> Then, the domain is R-{1}.

Now we will find the zeros.

==>** x = -12 **, **x= 1**, and **x = 2**

** However, x=1 is not...**

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((x+12)(x-2))/(x-1) >= 0

First we find the domain for the inequality.

==> The inequality if not defined when **x = 1**.

==> Then, the domain is R-{1}.

Now we will find the zeros.

==>** x = -12 **, **x= 1**, and **x = 2**

**However, x=1 is not in the domain so we will use open interval.**

==> Then we have the following intervals:

==> `(-oo, -12] , [-12,1), (1, 2] and [2, oo)`

`` ==> `(-oo, -12``] ` ==> x = -13 ==> -1*-15/-14 < 0

==> [-12, 1) ==> x = 0 ==> 12*-2/-1 > 0

==> (1, 2] ==> x = 3/2 ==> 13.5*-0.5/0.5 < 0

==>`[2, oo)` ==> x = 3 ==> 15*1/2 > 0

Then the inequality holds true for the following interval:

`x in [-12, 1) U [2, oo)`