2x³ + x² - 12x + 9 < 0

We begin by trying to factor.  A good way to start is
by dividing it by x-1 or by x+1, because 1 is always 
a possible zero or root of a polynomial.  It may not work
but it's always a good try.

1 | 2  1 -12  9
  |    2   3 -9  
    2  3  -9  0

Hot dog!  It worked!  So we have now factored
it as

(x - 1)(2x² + x - 12) < 0

And now we can factor the quadratic:

(x - 1)(2x - 3)(x + 3) < 0

So there are three critical values, found by
setting all factors = 0

x = 1; x = ); x = -3

Mark those on a number line:

-------------o-----------o-o------------------
-6  -5  -4  -3  -1   0   1   2   3   4   5   6

They divide the number line into 4 intervals:

(-∞,-3), (-3,1), (1,), (,∞)

For any of these intervals between or beyond critical 
values, if (x - 1)(2x - 3)(x + 3) < 0 is true (respectively, 
false) for one value in that interval, then it is true 
(respectively, false) for all values in that interval, so we 
take a test value for x in each interval.

For the interval (-∞,-3), the easiest test point is -4:

(x - 1)(2x - 3)(x + 3) < 0

(-4 - 1)(2(-4) - 3)(-4 + 3) < 0
             (-5)(-8-3)(-1) < 0
              (-5)(-11)(-1) < 0
                        -55 < 0

That is true, so (-∞,-3) is part of the solution:

<============o-----------o-o------------------
-6  -5  -4  -3  -1   0   1   2   3   4   5   6



For the interval (-3,1), the easiest test point is 0:

(x - 1)(2x - 3)(x + 3) < 0

(0 - 1)(2(0) - 3)(0 + 3) < 0
            (-1)(0-3)(3) < 0
             (-1)(-3)(3) < 0
                       3 < 0

That is false, so (-3,1) is not part of the solution:


For the interval (1,), , the easiest test point is 1.1:

(x - 1)(2x - 3)(x + 3) < 0

(1.1 - 1)(2(1.1) - 3)(-1.1 + 3) < 0
               (0.1)(2.2-3)(-1) < 0
               (0.1)(-0.8)(1.9) < 0
                          -.152 < 0

That is true, so (1,)  is part of the solution:

<============o-----------o=o------------------
-6  -5  -4  -3  -1   0   1   2   3   4   5   6

For the interval (, ∞), the easiest test point is 2:

  (x - 1)(2x - 3)(x + 3) < 0

(2 - 1)(2(2) - 3)(2 + 3) < 0
             (1)(4-3)(5) < 0
               (1)(1)(5) < 0
                       5 < 0

That is false, so (, ∞) is not part of the solution:

<============o-----------o=o------------------
-6  -5  -4  -3  -1   0   1   2   3   4   5   6

So the solution in interval notation is

(-∞,-3) U (1,)

What the problem is asking is: Where is the graph of 
y = 2x³ + x² - 12x + 9 below the x-axis?:



And we see that it is below the x-axis (less than 0)
to the left of -3, and between 1 and . 

Edwin