Determine the solution set of (x+1)(x-2)/x+3 >= 0

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`((x+1)(x-2))/(x+3) gt= 0`

The sign of the inequality can be determined by both numerator and denominator. Therefore I would rearrange the expression as below.

`((x+1)(x-2))/(x+3) xx (x+3)/(x+3)gt= 0`

`((x+1)(x-2)(x+3))/(x+3)^2 gt= 0`

Now the denominator is always positive and so we can evaluate the numerator for the inequality.

`(x+1)(x-2)(x+3) >= 0`

We can indentify the ranges for this inequaility as below.

`xlt=-3,`

Then, `(x+1)(x-2)(x+3) <= 0`

`-3lt=xlt-1`

Then, `(x+1)(x-2)(x+3) >= 0`

`-1lt=xlt=2`

Then, `(x+1)(x-2)(x+3) <= 0`

`xgt= 2`

Then, `(x+1)(x-2)(x+3) >= 0`

**Therefore the solutions for the given inequality are** `-3lt=xlt-1` **and** `xgt= 2.`

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