Question 1046372
(x+3)(x-2) / (x+2)(x-1) >_ 0
My guess is that what you wrote is not the inequality which you really have.


Maybe you really have  ((x+3)(x-2))/((x+2)(x-1)) >=0
and the rendering makes to appear as
{{{((x+3)(x-2))/((x+2)(x-1))>=0}}}.



The critical values of x are those at which the sign of a factor changes.  The critical x values here are  -3, -2, 2, 1.
These values cut the real number line into the five intervals:
{{{system(-infinity<x<=-3,-3<=x<=-2,-2<=x<=1,1<=x<=2,2<=x<infinity)}}}.



What you do is test ANY value in each interval to find if it makes the expression GREATER THAN OR EQUAL TO 0.  Best done as watching the sign of each factor of the expression.

<pre> 

                   Pick Value         SIGNS           RESULT SIGN        MEANS

(-inf,-3]           -4             (- -)/(- -)           +               TRUE

[-3,-2]             -5/2          (+ -)/(- -)            -               FALSE

[-2,1]              -1            (+ -)/(+ -)            +               TRUE

[1,2]               +3/2          (+ -)/(+ +)            -               FALSE

[2,infin)           +3            (+ +)/(+ +)            +               TRUE
</pre>


Note that "0" would be just as good for TRUTH as "POSITIVE", because the given inequality is for the expression GREATER than or EQUAL to zero.  Positive means greater than zero.