To determine To calculate: The solution of the inequality ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) > 0 in the interval notation. Explanation Given Information: The provided inequality is ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) > 0 . Formula used: Follow the steps as mentioned below to solve the quadratic, polynomial and rational inequalities. Step1: Compute the boundary values by solving the inequality for x via treating inequality as equation. Step2: Draw a number line for the solution obtained from step 1 by using following points, Use open circle ∘ for the inequality < or > . Use closed circle • for the inequality ≤ or ≥ . Step3: For a rational inequality compute the boundary values by equating the denominator equal to zero, represent these boundary conditions on number line by open circle ∘ . Step4: From each interval check the inequality at any test point within the interval. Step5: The set of points that satisfy the inequality will be the solution. Calculation: Consider the inequality, ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) > 0 Apply zero factor property as, Equate 4 x − 1 to 0 and then solve for x , 4 x − 1 = 0 4 x = 1 x = 1 4 Equate x + 3 to 0 and then solve for x , x + 3 = 0 x = − 3 Equate x − 2 to 0 and then solve for x , x − 2 = 0 x = 2 Therefore, the boundary values for the inequality are − 3 , 1 4 and 2 . Here the inequality is “ > ”, so, use open circle to represent the solution and the boundary values on the number line, as, Form the table to check the test points with in the interval for all the possible intervals, Interval A ( − ∞ , − 3 ) Test value − 4 ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) > 0 ( 4 ( − 4 ) − 1 ) ( − 4 + 3 ) ( − 4 − 2 ) > ? 0 ( − 17 ) ( − 1 ) ( − 6 ) > ? 0 − 102 > 0 This is not possible, so, this interval is false

Intermediate Algebra For College Students (10th Edition)

10th Edition

ISBN: 9780134758992

Author: Allen R. Angel, Dennis Runde

Publisher: PEARSON

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Chapter 8.6, Problem 25ES

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To calculate: The solution of the inequality (4x−1)(x+3)(x−2)>0 in the interval notation.

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Chapter 8.6, Problem 24ES

Chapter 8.6, Problem 26ES

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Intermediate Algebra For College Students (10th Edition)

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The solution of the inequality ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) &gt; 0 in the interval notation.

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To calculate: The solution of the inequality (4x−1)(x+3)(x−2)>0 in the interval notation.

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Chapter 8 Solutions

The solution of the inequality ( 4 x − 1 ) ( x + 3 ) ( x − 2 ) > 0 in the interval notation.