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 equals 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: Consider polynomial inequality x 2 + 4 x + 7 ≥ 5 x + 9 . Rewrite the polynomial inequality as follows, x 2 + 4 x + 7 − 5 x − 9 ≥ 0 x 2 − x − 2 ≥ 0 By the procedure from 1 to 4, find the solution set of the inequality. Change the inequality sign by equal sign to form equivalent equation as x 2 − x − 2 = 0 . In order to solve the equation, first factor it by grouping as follows. x 2 − x − 2 = 0 x 2 − 2 x + x − 2 = 0 x ( x − 2 ) + 1 ( x + 2 ) = 0 ( x + 1 ) ( x − 2 ) = 0 By zero product principle, find the value of x as follows. x + 1 = 0 x = − 1 x − 2 = 0 x = 2 The solution set is { − 1 , 2 } . The solutions divide the x axis into the intervals as ( − ∞ , − 1 ) , ( − 1 , 2 ) and ( 2 , ∞ ) . Let the polynomial equation f ( x ) = x 2 − x − 2 . Find the sign of the polynomial on each interval by selecting test value as follows. Obtain the sign of polynomial f ( x ) = x 2 − x − 2 at x = − 3 on the interval ( − ∞ , − 1 )

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ISBN: 2810000027802

Author: BITTINGER

Publisher: PEARSON

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Chapter 4.6, Problem 30E

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The polynomial inequality x 2 + 4 x + 7 ≥ 5 x + 9 .

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 x2+4x+7≥5x+9.

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