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.