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Write the equation of a real function with zeroes at x = 1 and x = -2+5i. The imaginary unit í should not be part of the answer. f(x) =

Write the equation of a real function with zeroes at x = 1 and x = -2+5i. The imaginary unit í should not be part of the answer. f(x) =

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Title: Writing a Real Function with Given Zeroes --- **Objective:** Learn how to construct a real function given specific zeroes. --- **Problem Statement:** Write the equation of a real function with zeroes at \(x = 1\) and \(x = -2 + 5i\). **Note:** The imaginary unit \(i\) should not be part of the answer. **Solution:** To find the equation, consider the following steps: 1. **Real Zeroes:** - \(x = 1\) can be written as \((x - 1)\). 2. **Complex Conjugates:** - Since complex roots occur in conjugate pairs for real functions, the zero \(x = -2 + 5i\) implies there is also a zero at \(x = -2 - 5i\). - These can be written as \((x + 2 - 5i)\) and \((x + 2 + 5i)\). 3. **Construct the Function:** - The polynomial can be constructed by multiplying these factors: \[ f(x) = (x - 1)(x + 2 - 5i)(x + 2 + 5i) \] - To remove the imaginary unit, multiply the conjugates: \[ (x + 2 - 5i)(x + 2 + 5i) = (x + 2)^2 - (5i)^2 = (x + 2)^2 + 25 \] - Therefore, the function is: \[ f(x) = (x - 1)((x + 2)^2 + 25) \] **Final Function:** - Expand if necessary to find the complete polynomial expression: \[ f(x) = (x - 1)(x^2 + 4x + 4 + 25) = (x - 1)(x^2 + 4x + 29) \] This represents a real polynomial with the specified zeroes. ---
Transcribed Image Text:Title: Writing a Real Function with Given Zeroes --- **Objective:** Learn how to construct a real function given specific zeroes. --- **Problem Statement:** Write the equation of a real function with zeroes at \(x = 1\) and \(x = -2 + 5i\). **Note:** The imaginary unit \(i\) should not be part of the answer. **Solution:** To find the equation, consider the following steps: 1. **Real Zeroes:** - \(x = 1\) can be written as \((x - 1)\). 2. **Complex Conjugates:** - Since complex roots occur in conjugate pairs for real functions, the zero \(x = -2 + 5i\) implies there is also a zero at \(x = -2 - 5i\). - These can be written as \((x + 2 - 5i)\) and \((x + 2 + 5i)\). 3. **Construct the Function:** - The polynomial can be constructed by multiplying these factors: \[ f(x) = (x - 1)(x + 2 - 5i)(x + 2 + 5i) \] - To remove the imaginary unit, multiply the conjugates: \[ (x + 2 - 5i)(x + 2 + 5i) = (x + 2)^2 - (5i)^2 = (x + 2)^2 + 25 \] - Therefore, the function is: \[ f(x) = (x - 1)((x + 2)^2 + 25) \] **Final Function:** - Expand if necessary to find the complete polynomial expression: \[ f(x) = (x - 1)(x^2 + 4x + 4 + 25) = (x - 1)(x^2 + 4x + 29) \] This represents a real polynomial with the specified zeroes. ---
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