SOLUTION: (x+18)(x-2)(x+11)>0 I need to use at least one inequality or compound inequality to express the answer in the form of a solution set. I cannot figure out how to do this! I thank

Algebra ->  Inequalities -> SOLUTION: (x+18)(x-2)(x+11)>0 I need to use at least one inequality or compound inequality to express the answer in the form of a solution set. I cannot figure out how to do this! I thank      Log On


   

Question 147153: (x+18)(x-2)(x+11)>0
I need to use at least one inequality or compound inequality to express the answer in the form of a solution set. I cannot figure out how to do this!
I thank you so much for any help!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B11%29%28x-2%29%28x%2B18%29%3E0 Start with the given inequality


%28x%2B11%29%28x-2%29%28x%2B18%29=0 Set the left side equal to zero


Set each individual factor equal to zero:

x%2B11=0, x-2=0 or x%2B18=0

Solve for x in each case:

x=-11, x=2 or x=-18


So our critical values are x=-11, x=2 and x=-18

Now set up a number line and plot the critical values on the number line

number_line%28+600%2C+-20%2C+20%2C-11%2C2%2C-18%29



So let's pick some test points that are near the critical values and evaluate them.


Let's pick a test value that is less than -18 (notice how it's to the left of the leftmost endpoint):

So let's pick x=-19

x%5E3%2B27x%5E2%2B140x-396%3E0 Start with the given inequality


%28-19%2B11%29%28-19-2%29%28-19%2B18%29%3E+0 Plug in x=-19


-168%3E+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.


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Let's pick a test value that is in between -18 and -11:

So let's pick x=-14

x%5E3%2B27x%5E2%2B140x-396%3E0 Start with the given inequality


%28-14%2B11%29%28-14-2%29%28-14%2B18%29%3E+0 Plug in x=-14


192%3E+0 Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is ()





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Let's pick a test value that is in between -11 and 2:

So let's pick x=-4

x%5E3%2B27x%5E2%2B140x-396%3E0 Start with the given inequality


%28-4%2B11%29%28-4-2%29%28-4%2B18%29%3E+0 Plug in x=-4


-588%3E+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.


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Let's pick a test value that is greater than 2 (notice how it's to the right of the rightmost endpoint):

So let's pick x=3

x%5E3%2B27x%5E2%2B140x-396%3E0 Start with the given inequality


%283%2B11%29%283-2%29%283%2B18%29%3E+0 Plug in x=3


294%3E+0 Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is ()





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Summary:

So the solution in interval notation is:


() ()





Here's a graph to prove it
+graph%28+500%2C+500%2C+-30%2C+10%2C+-630%2C+300%2C+%28x%2B11%29%28x-2%29%28x%2B18%29%29+ Graph of y=%28x%2B11%29%28x-2%29%28x%2B18%29