Explanation Calculation: The given equation is ( x + 3 ) ( x 2 − 4 ) = 0 . By inspection, substitute x = − 1 in the equation ( x + 3 ) ( x 2 − 4 ) = 0 . ( − 1 + 3 ) ( ( − 1 ) 2 − 4 ) = ? 0 ( 2 ) ( − 3 ) = ? 0 − 6 = 0 ( False ) Therefore, x = − 1 is not the root of the equation ( x + 3 ) ( x 2 − 4 ) = 0 . Substitute x = − 2 in the equation ( x + 3 ) ( x 2 − 4 ) = 0 . ( − 2 + 3 ) ( ( − 2 ) 2 − 4 ) = ? 0 ( 1 ) ( 4 − 4 ) = ? 0 0 = 0 ( True ) Therefore, x = − 2 is a root of the equation ( x + 3 ) ( x 2 − 4 ) = 0 . Substitute x = 2 in the equation ( x + 3 ) ( x 2 − 4 ) = 0

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 15.2, Problem 3E

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The roots of the equation (x+3)(x2−4)=0 using inspection.

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Chapter 15.2, Problem 2E

Chapter 15.2, Problem 4E

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Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2-t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = 1+t, y = 2 − t, z = 4 - 3t and is parallel to the plane 5x + 2y + z = 1. (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1+t, and z = 2-t. (e) The plane that contains the lines L₁: x = 1 + t, y = 1 − t, z = 2t and L2 : x = 2 − s, y = s, z = 2.

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Chapter 15 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 15.1 - Use the factor theorem to determine whether t − 2...Ch. 15.1 - Prob. 2PE Ch. 15.1 - Prob. 3PE Ch. 15.1 - Prob. 4PE Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - In Exercises 1–4, make the given changes in the...Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E Ch. 15.1 - Prob. 6E

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The roots of the equation ( x + 3 ) ( x 2 − 4 ) = 0 using inspection.

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Chapter 15 Solutions