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: The given polynomial inequality is f ( x ) ≤ 0 , where f ( x ) = x 2 + 2 x − 15 . By the above procedure, firstly obtain the solution set of the inequality. That is, obtain the solution of the equation x 2 + 2 x − 15 = 0 . x 2 + 2 x − 15 = 0 x 2 + 5 x − 3 x − 15 = 0 x ( x + 5 ) − 3 ( x + 5 ) = 0 ( x + 5 ) ( x − 3 ) = 0 By the above principle, the value of x as follows: ( x + 5 ) = 0 or ( x − 3 ) = 0 x = 3 or x = − 5 Therefore, the solution set of the polynomial equation x 2 + 2 x − 15 = 0 is { − 5 , 3 } . The solutions divide the x axis into the intervals as ( − ∞ , − 5 ) , ( − 5 , 3 ) and ( 3 , ∞ ) . Find the sign of the polynomial on each interval by selecting test value as follows. Obtain the sign of the polynomial f ( x ) = x 2 + 2 x − 15 at x = − 6 on the interval ( − ∞ , − 5 )

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ISBN: 9780134383385

Author: Penna

Publisher: VST

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Chapter 4.6, Problem 3E

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To solve: The polynomial inequality f(x)≤0, where f(x)=x2+2x−15.

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Chapter 4.6, Problem 2E

Chapter 4.6, Problem 4E

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The polynomial inequality f ( x ) ≤ 0 , where f ( x ) = x 2 + 2 x − 15 .

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To solve: The polynomial inequality f(x)≤0, where f(x)=x2+2x−15.

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Chapter 4 Solutions