# Solve Polynomial Inequalities - Tutorial

Solving polynomial inequalities: A step by step tutorial with examples and detailed solutions. A review on the sign of polynomial expressions is first presented.

Review:

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

x2 < -x + 6

Solution to Example 1:

• Given
x2 < -x + 6

• Rewrite the inequality with one side equal to zero.
x2 + 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.

 -3 2 + 0 - 0 +

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

x2 + 4x < 5

Example 2: Solve the polynomial inequality

(x2 + 2)(x + 1)(x + 6) > 0

Solution to Example 2:

• Given
(x2 + 2)(x + 1)(x + 6) > 0

• Let us note that (x2 + 2) is always positive and the sign of the polynomial depends on the zeros -6 and -1. The two real zeros -6 and -1 divide the real number line into 3 intervals.
(-∞ , -6)  (-6 , -1)  and  (-1 , +∞)

• 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 (x2 + 2)(x + 1)(x + 6).

• a) (-∞ , -6)

• chose x = - 7 and evaluate (x2 + 2)(x + 1)(x + 6)

(x2 + 2)(x + 1)(x + 6) = ((-7)2 + 2)(-7 + 1)(-7 + 6)

= 51*(-6)*(-1)

= 306

(x2 + 2)(x + 1)(x + 6) is positive in (-∞ , -1)

• b) (-6 , -1)

• chose x = - 2 and evaluate (x2 + 2)(x + 1)(x + 6)

(x2 + 2)(x + 1)(x + 6) = ((-2)2 + 2)(-2 + 1)(-2 + 6)

= 6*(-1)*4

= - 24

(x2 + 2)(x + 1)(x + 6) is negative in (-6 , -1)

• c) (-1 , +∞)

• chose x = 0 and evaluate (x2 + 2)(x + 1)(x + 6)

(x2 + 2)(x + 1)(x + 6) = ((0)2 + 2)(0 + 1)(0 + 6)

= 12

(x2 + 2)(x + 1)(x + 6) is positive in (-1 , +∞)

We now put all the information obtained above in a table.

 -6 -1 + 0 - 0 +

Conclusion

The solution set consists of all real numbers in (- ∞ , -6) U (-1 , + ∞).

Matched Exercise: Solve the polynomial inequality

-(x6 + 8)(x - 1)(x + 4) < 0

More references and links on how to Solve Equations, Systems of Equations and Inequalities.