easy way to do these ....
Since the expression is already factored, you are off to a good start.
the critical values are x = -5,-2 and 2
(those would have been the solution had it been an equation)
so we consider 4 sections:
1. x < -5
take any value in that domain, say x = -10
we get (-)(-)(-) < 0 ----- false
2. -5 < x < -1
take any value in that domain, say x = -4
(+)(-)(-) > 0 ----- true
3. -1 < x < 2
take any value of x in that domain, say x = 0
(+)(-)(+) < 0 ---- false
4. x > 2
let's pick x = 10
(+)(+)(+) > 0 ------ true
so solution:
-5 ≤ x ≤ -1 OR x > 2
(x+5)(x-2)(x+1)>0
What is the solution set? Why am I not getting this at all?
2 answers
third line contains a typo, should have said ...
the critical values are x = -5,-1 and 2
the critical values are x = -5,-1 and 2