SOLUTION: solve (x+3)(x-16)(x+2)>0 The solution set is {x ⎹ }

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Question 365114: solve (x+3)(x-16)(x+2)>0 The solution set is {x ⎹ }
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Break up the number line into 4 regions using the critical points of the function.
Region 1:(-infinity,-3)
Region 2:(-3,-2)
Region 3:(-2,16)
Region 4:(16,infinity)
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For each region, choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution region.
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Region 1:x=-4
%28x%2B3%29%28x-16%29%28x%2B2%29%3E0
%28-4%2B3%29%28-4-16%29%28-4%2B2%29%3E0
%28-1%29%28-20%29%28-2%29%3E0+
-40%3E0+
False, Region 1 is not part of the solution region.
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Region 2:x=-5%2F2
%28x%2B3%29%28x-16%29%28x%2B2%29%3E0
%28-5%2F2%2B3%29%28-5%2F2-16%29%28-5%2F2%2B2%29%3E0
%281%29%28-37%2F2%29%28-1%2F2%29%3E0+
37%2F4%3E0+
True, Region 2 is part of the solution region.
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Region 3:x=0
%28x%2B3%29%28x-16%29%28x%2B2%29%3E0
%28-4%2B3%29%28-4-16%29%28-4%2B2%29%3E0
%283%29%28-16%29%282%29%3E0+
-96%3E0+
False, Region 3 is not part of the solution region.
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Region 4:x=17
%28x%2B3%29%28x-16%29%28x%2B2%29%3E0
%2817%2B3%29%2817-16%29%2817%2B2%29%3E0
%2820%29%281%29%2819%29%3E0+
380%3E0+
True, Region 4 is part of the solution region.
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Solution Region:(-3,-2) U (16,infinity)
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Graphical verification: Look for regions where the function is above the x-axis (y%3E0)
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