Find the critical values by
setting each factor = 0

 gives critical value 
 gives critical value 

Place those critical values on a number line:

-----------o-----------------------------------o---------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11 12 

Choose a test value for x in the region left of -3.
The easiest number to choose is -4.  Substitute this
value into either the original inequality or its 
factored form. I'll choose to substitute in the
factored form, but it does not matter which you
substitute  in:






This is false so we do not shade the region left of -3.

So we still have only:

-----------o-----------------------------------o---------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11 12


Choose a test value for x in the region right of -3 but
left of 9. The easiest number to choose is 0.  Substitute 
this value into either the original inequality or its 
factored form. I'll choose to substitute in the
factored form, but it does not matter which you
substitute  in:






This is true so we shade the region right of -3 but left of 9.

-----------o===================================o---------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11 12


Choose a test value for x in the region right of 9.
The easiest number to choose is 10.  Substitute this
value into either the original inequality or its 
factored form. I'll choose to substitute in the
factored form, but it does not matter which you
substitute  in:






This is false so we do not shade the region right of 9.

So we still have:

-----------o===================================o---------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11 12

Since the inequality is <, rather than <, neither
critical value is itself a solution.

So the final number line graph is:

-----------o===================================o---------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11 12
 
and the interval notation for that is (,).

Edwin