SOLUTION: I sent this problem earlier but after receiving a response quickly realized I entered the wrong problem, can u please help me with this one? I have no clue where to start. Thanks i

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I sent this problem earlier but after receiving a response quickly realized I entered the wrong problem, can u please help me with this one? I have no clue where to start. Thanks i      Log On


   

Question 242760: I sent this problem earlier but after receiving a response quickly realized I entered the wrong problem, can u please help me with this one? I have no clue where to start. Thanks in advance!!
Solve:
(x+2)(x-2)(x+1)>0
a)the solution set is
b)solution is all real numbers
c)there is no solution
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-2%29%28x%2B2%29%28x%2B1%29%3E0 Start with the given inequality.


%28x-2%29%28x%2B2%29%28x%2B1%29=0 Set the left side equal to zero


Set each individual factor equal to zero:

x-2=0, x%2B2=0 or x%2B1=0

Solve for x in each case:

x=2, x=-2 or x=-1


So our critical values are x=2, x=-2 and x=-1

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

number_line%28+600%2C+-10%2C+10%2C2%2C-2%2C-1%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 -2 (notice how it's to the left of the leftmost endpoint). This corresponds to the left most interval ()


So let's pick x=-3

%28x-2%29%28x%2B2%29%28x%2B1%29%3E0 Start with the given inequality


%28-3-2%29%28-3%2B2%29%28-3%2B1%29%3E+0 Plug in x=-3


-10%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 -2 and -1. This corresponds to the left middle interval ()

So let's pick x=-1.5

%28x-2%29%28x%2B2%29%28x%2B1%29%3E0 Start with the given inequality


%28-1.5-2%29%28-1.5%2B2%29%28-1.5%2B1%29%3E+0 Plug in x=-1.5


0.875%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 -1 and 2. This corresponds to the right middle interval ()

So let's pick x=0

%28x-2%29%28x%2B2%29%28x%2B1%29%3E0 Start with the given inequality


%280-2%29%280%2B2%29%280%2B1%29%3E+0 Plug in x=0


-4%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). This corresponds to the right most interval ()

So let's pick x=3

x%5E3%2Bx%5E2-4x-4%3E0 Start with the given inequality


%283-2%29%283%2B2%29%283%2B1%29%3E+0 Plug in x=3


20%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:


() ()

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


First thing to do is to solve the problem that you posted before, namely:



Take the three roots of this equation, call them , , and (in ascending numerical order) and create four intervals.









For each of the four intervals, select a value and substitute it back into the original inequality. Once you have done that, two of the values will result in true statements and two of them will make the original inequality false. The union of the two intervals from which the values that made the original inequality true is the solution set of your inequality.

John