SOLUTION: hello I do not have an exact idea on how to solve this problem below (x+2)(x-14)(x+1)>0 There will either be a solution set for {x| } or it will be all real numbers

Question 249088: hello
I do not have an exact idea on how to solve this problem below
(x+2)(x-14)(x+1)>0

There will either be a solution set for {x| }
or it will be all real numbers or finally, no solution???
I got as far as multiplying them threw the distributive process. Dont know if this was necessary or not..???
x3rd power -11x2nd power-40x-28 > 0

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I do not have an exact idea on how to solve this problem below
(x+2)(x-14)(x+1)>0
-----------------
1st: Since the left side cannot be zero
x cannot be -2,14,or -1
---------------------------------
2nd: Draw a number line and mark -2,-1,14 on it.
---------------------------------
3rd: That breaks the line up into 4 intervals.
Choose a check point in each interval to see where
the solutions for the INEQUALITY are.
Test each check point in (x+2)(x-14)(x+1)>0
Note: You only need to check the sign of the answer because you
are comparing the value to zero.
---------------------------------------
Checking:
If x = -3, you get -*-*- > 0 ; false, so no solutions in (-inf,-2)
--------
If x = -3/2, you get +*-*- > 0 ; true, so solutions in (-2,-1)
--------
If x = 13, you get +*-*+ > 0 ; false, so no solutions in (-1,14)
--------
If x = 15, you get +*+*+ > 0 ; true, so solutions in (14,+inf)
--------------------
Final Answer: (-2,-1)U(14,+inf)
======================================
Cheers,
Stan H.
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