SOLUTION: Example 4 on page 352 shows how to do this problem, but I am having problems factoring once I get to the point of setting the 2-x-x^2>0 and then solving for x (factoring is one of

Question 151505This question is from textbook precalculus
: Example 4 on page 352 shows how to do this problem, but I am having problems factoring once I get to the point of setting the 2-x-x^2>0 and then solving for x (factoring is one of my major weaknesses). I seem to have trouble with factoring whenever I am faced with needing to combine the same variable in two different forms (e.g. the -x-x^2). Thanks. This question is from textbook precalculus

Found 2 solutions by checkley77, jim_thompson5910:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
2-x-x^2>0
-X^2-X+2>0
(-X+1)(X+2)=0
-X+1=0
-X=-1
X=1 ANSWER.
X+2=0
X=-2 ANSWER.

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given inequality.

Rearrange the terms.

Rearrange the terms.

Factor the left side (note: if you need help with factoring, check out this solver)

Set the left side equal to zero

Set each individual factor equal to zero:

or

Solve for x in each case:

or

So our critical values are and

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

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 (notice how it's to the left of the leftmost endpoint):

So the interval that we are testing is ()

So let's pick to test the interval

Start with the given inequality

Plug in

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.

---------------------------------------------------------------------------------------------

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

So the interval that we are testing is ()

So let's pick to test the interval

Start with the given inequality

Plug in

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 ()

---------------------------------------------------------------------------------------------

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

So the interval that we are testing is ()

So let's pick

Start with the given inequality

Plug in

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.

---------------------------------------------------------------------------------------------

Summary:

So the solution in interval notation is:

()

Here's a graph to visually verify the answer:

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