A polynomial can change sign only at its real zeros. When ordered, the real zeros of a
polynomial divide the real number line into
intervals in which the polynomial does not change sign.
Example 1: Solve the polynomial inequality
x^{2} < x + 6
Solution to Example 1:

Given
x^{2} < x + 6

Rewrite the inequality with one side equal
to zero.
x^{2} + x  6 < 0

Factor the left side of the inequality.
(x  2)(x + 3) < 0

The two real zeros 3 and 2 of the left side of the inequality, divide the
real number line into 3 intervals.
(∞ , 3) (3 , 2) and (2 , +∞)

The sign within each interval is
determined by using test values. We chose one value within each interval and use
it to find the sign of (x  2)(x + 3).
 a) (∞ , 3)

chose x = 4 and evaluate (x  2)(x + 3)
(x  2)(x + 3) = (4  2)(4 + 3)
= 6
(x  2)(x + 3) is positive in (∞ ,
3)
 b) (3 , 2)

chose x = 0 and evaluate (x  2)(x + 3)
(x  2)(x + 3) = (0  2)(0 + 3)
= 6
(x  2)(x + 3) is negative in (3 , 2)
 c) (2 , +∞)

chose x = 4 and evaluate (x  2)(x + 3)
(x  2)(x + 3) = (4  2)(4 + 3)
= 14
(x  2)(x + 3) is positive in (2 , +∞)

We now put all the above information in a
table.
Conclusion
We are looking for values of x that make (x  2)(x + 3)
negative. The solution set consists of all real numbers in the
interval (3 , 2).
Matched Exercise:Solve the polynomial inequality
x^{2} + 4x < 5
Example 2: Solve the polynomial inequality
(x^{2} + 2)(x + 1)(x + 6) > 0
Solution to Example 2:
Conclusion
The solution set consists of all real numbers in ( ∞ , 6) U (1 , + ∞).
Matched Exercise: Solve the polynomial inequality
(x^{6} + 8)(x  1)(x + 4) < 0
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