Explanation Principle used: If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. That is If A B = 0 , then A = 0 or B = 0 . Procedure used: 1 . Find an equivalent inequality with P ( x ) on one side and 0 on the other. 2. Change the inequality symbol to an equal sign and solve the related equation; that is, solve P ( x ) = 0 . 3. Use the solutions to divide the x axis into intervals. Then select a test value from each interval and determine the sign of the polynomial on the interval. 4. determine the intervals for which the inequality is satisfied and write intervals notation or set builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is ≤ or ≥ . Calculation: The given polynomial inequality is x 5 + x 2 ≥ 2 x 3 + 2 . By the procedure mentioned above, find the solution set of the inequality. An equivalent inequality with P ( x ) on one side and 0 on the other is x 5 − 2 x 3 + x 2 − 2 ≥ 0 Change the inequality sign by an equal sign to form the equivalent equation as x 5 − 2 x 3 + x 2 − 2 = 0 . In order to solve the equation, first factor it by groupingas follows. x 5 − 2 x 3 + x 2 − 2 = 0 ( x 3 + 1 ) ( x 2 − 2 ) = 0 x 3 + 1 = 0 or x 2 − 2 = 0 x = − 1 or x = ± 2 The solution set of the equation is { − 2 , − 1 , 2 } . The solutions divides the x axis into the intervals as ( − ∞ , − 2 ) , ( − 2 , − 1 ) , ( − 1 , 2 ) and ( 2 , ∞ ) . Find the sign of the polynomial on each interval by selecting test values as follows. Let the polynomial function be f ( x ) = x 5 − 2 x 3 + x 2 − 2 . Obtain the sign of polynomial f ( x ) at x = − 2 on the interval ( − ∞ , − 2 ) . f ( − 2 ) = ( − 2 ) 5 − 2 ( − 2 ) 3 + ( − 2 ) 2 − 2 = − 32 + 16 + 4 − 2 = − 14 < 0 Obtain the sign of polynomial f ( x ) at x = − 1

College Algebra: Graphs and Models (6th Edition)

6th Edition

ISBN: 9780134179032

Author: Marvin L. Bittinger, Judith A. Beecher, David J. Ellenbogen, Judith A. Penna

Publisher: PEARSON

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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

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Chapter 4.6, Problem 45E

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To solve: The polynomial inequality x5+x2≥2x3+2.

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Chapter 4.6, Problem 44E

Chapter 4.6, Problem 46E

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Chapter 4 Solutions

College Algebra: Graphs and Models (6th Edition)

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The polynomial inequality x 5 + x 2 ≥ 2 x 3 + 2 .

Solution Summary: The author explains how to solve the polynomial inequality by dividing the x axis into intervals.

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To solve: The polynomial inequality x5+x2≥2x3+2.

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Chapter 4 Solutions