Solve (x+5)(x-13)(x+4) > 0?

Question

They want us to "Use at least one inequality or compound inequality to express your answer. For answers with more than one inequality, separate the inequalities by a comma or the word 'or'. Type R if the answer is all real #s. Type N if there is no real solution."

I do not have a clue on how to do this. I have tried and tried, can someone please help me?

Answer 1

x + 5 > 0

x > - 5

x - 13 > 0

x > 13

x + 4 > 0

x > - 4

(x + 5)(x - 13)(x + 4) > 0

(x² - 13x + 5x - 65)(x + 4) > 0

(x² - 8x - 65)(x + 4) > 0

x³ - 8x² - 65x + 4x² - 32x - 260 > 0

x³ - 4x² - 97x > 260

Answer 2

Just treat all the parens as seperate questions.

(x+5) > 0

(x-13) > 0

(x+4) > 0

and solve individually.

x > -5

x > 13

x > -4

since it's multiplication and if you have even one neg #, the result will be < 0, x must be > -4

Answer 3

Consider A = (x+5)(x-13)(x+4) as A = P*Q*R

So A is positive (>0) in the following conditions:

1. P>0, Q>0, R>0

2. P>0, Q<0, R<0

3. P<0, Q>0, R<0

4. P<0, Q<0, R>0

Now solve them separately and get values of x