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
