Prentice Hall Algebra 2

Chapter 5 Polynomials and Polynomial Functions > Concept Byte: Solving Polynomial Inequalities

Concept Byte: Solving Polynomial Inequalities

Use With Lesson 5-5

EXTENSION

You can use what you know about factoring and graphing polynomials to help you solve polynomial inequalities.

Example 1

Solve 12 x 3 + 11 x 2 − 7 x − 6 < 0 . 12 x cubed , plus , 11 x squared , minus 7 x minus 6 less than 0 .

By the Rational Root Theorem and the Factor Theorem, open x plus 1 close is one factor.

So , 12 x 3 + 11 x 2 − 7 x − 6 = ( x + 1 ) ( 12 x 2 − x − 6 ) = ( x + 1 ) ( 3 x + 2 ) ( 4 x − 3 ) table with 2 rows and 2 columns , row1 column 1 , cap so comma 12 , x cubed , plus 11 , x squared , minus 7 x minus 6 , column 2 equals . open , x plus 1 , close . open . 12 , x squared , minus x minus 6 . close , row2 column 1 , , column 2 equals . open , x plus 1 , close . open , 3 x plus 2 , close . open , 4 x minus 3 , close , end table

The zeros of the polynomial are − 1 , − 2 3 , negative 1 comma negative , 2 thirds , comma and 3 fourths , .

You know the graph of a cubic polynomial with a positive leading coefficient has the shape shown below. The polynomial is less than 0 where the graph is below the x-axis.

An N-shaped graph rises through (negative 1, 0) to a vertex at (negative seven-eighths), falls through (0, negative 7) to a vertex at (one-quarter, negative 7 and a half), and rises through (3 over 4, 0). All values are approximate.

Thus, x less than negative 1 and − 2 3 < x < 3 4 negative , 2 thirds , less than x less than , 3 fourths is the solution of the inequality.

Example 2

Solve − x 3 + 12 x 2 − 44 x + 54 > 0 . negative , x cubed , plus , 12 x squared , minus 44 x plus 54 greater than 0 .

You know the shape of the graph. Use a graphing calculator to find the only zero at x equals 6 . 53 .

A graphing calculator screen: a graph falls away from the positive y-axis to a vertex at (3, 1), rises to a vertex at (5, 2), and then falls through the x-axis at (6.5251023, 0).

The graph is above the x-axis when x less than 6 . 53 . Therefore, the solution of the inequality is x less than 6 . 53 .

Exercises

Solve each inequality.

8 x 2 − 2 x − 21 > 0 8 x squared , minus 2 x minus 21 greater than 0
− 16 x 2 + 104 x − 169 < 0 negative 16 , x squared , plus 104 x minus 169 less than 0
x 3 − 20 x 2 − 100 x + 2000 > 0 x cubed , minus , 20 x squared , minus 100 x plus , 2000 , greater than 0
− 10 x 3 + x 2 − 10 x + 1 > 0 negative 10 , x cubed , plus , x squared , minus 10 x plus 1 greater than 0
x ( x 2 + 1 ) > 2 ( 2 x 2 − 3 ) x open , x squared , plus 1 close greater than 2 open 2 , x squared , minus 3 close
7 x 3 ≤ x 2 + 66 x 7 x cubed , less than or equal to , x squared , plus 66 x
x 3 − 11 x 2 + 32 x − 28 ≥ 0 x cubed , minus , 11 x squared , plus 32 x minus 28 greater than or equal to 0
11 x 3 − 76 x 2 + 80 x + 192 > 0 11 x cubed , minus , 76 x squared , plus 80 x plus 192 greater than 0
Reasoning If Q(x) is a quartic polynomial, one possible solution of q open x close less than 0 is an interval of numbers, not including endpoints. Describe other possible solutions.

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Concept Byte: Solving Polynomial Inequalities

Example 1

Example 2

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