SOLUTION: Solve the inequality (x-2)^2(x+1)^3(x-5) <_0 if you could please thoroughly explain it because I am completely lost on this one. I tried to distribute (not sure if we're even su

Algebra ->  Inequalities -> SOLUTION: Solve the inequality (x-2)^2(x+1)^3(x-5) <_0 if you could please thoroughly explain it because I am completely lost on this one. I tried to distribute (not sure if we're even su      Log On


   



Question 1090338: Solve the inequality (x-2)^2(x+1)^3(x-5) <_0
if you could please thoroughly explain it because I am completely lost on this one. I tried to distribute (not sure if we're even supposed to) the entire thing and got lost

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
No, break up the number line into distinct regions using the critical values of the inequality (values that make the left hand side equal zero).
So,
x=2
x=-1
x=5
So then the four regions are:
Region 1: x%3C-1
Region 2: -1%3Cx%3C2
Region 3: 2%3Cx%3C5
Region 4: x%3E5
For each region, choose a value in the region (not an endpoint), test the inequality, determine if that point is part of the solution.
Region 1: x%3C-1
x=-2
%28x-2%29%5E2%2A%28x%2B1%29%5E3%2A%28x-5%29%3C=0
%28-2-2%29%5E2%2A%28-2%2B1%29%5E3%2A%28-2-5%29%3C=0
%28-4%29%5E2%2A%28-1%29%5E3%2A%28-7%29%3C=0
%2816%29%2A%28-1%29%2A%28-7%29%3C=0
112%3C=0
False, not part of the solution.
Region 2: -1%3Cx%3C2
x=0
%28x-2%29%5E2%2A%28x%2B1%29%5E3%2A%28x-5%29%3C=0
%280-2%29%5E2%2A%280%2B1%29%5E3%2A%280-5%29%3C=0
%284%29%2A%281%29%2A%28-5%29%3C=0
-5%3C=0
True, part of the solution.
Region 3: 2%3Cx%3C5
x=3
%28x-2%29%5E2%2A%28x%2B1%29%5E3%2A%28x-5%29%3C=0
%283-2%29%5E2%2A%283%2B1%29%5E3%2A%283-5%29%3C=0
%281%29%2A%2864%29%2A%28-2%29%3C=0
-128%3C=0
True, part of the solution.
Region 4: x%3E5
x=6
%28x-2%29%5E2%2A%28x%2B1%29%5E3%2A%28x-5%29%3C=0
%286-2%29%5E2%2A%286%2B1%29%5E3%2A%286-5%29%3C=0
%2816%29%2A%28343%29%2A%281%29%3C=0
5488%3C=0
False, not part of the solution.
So then, the solution region is
-1%3Cx%3C2U2%3Cx%3C5
The inequality also holds when x=2 because of the equal sign in the inequality so,
-1%3C=x%3C=5
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