Explanation Formula used: 1) z = z ¯ , if and only if z ∈ ℝ 2) Theorem: Conjugate Zeros: Suppose that f ( x ) is a polynomial that has all its coefficients in the real numbers. If the complex number z is a zero of f ( x ) , then z ¯ is also a zero of f ( x ) . Proof: Let f ( x ) be a polynomial of odd degree with real coefficients. That is, deg f ( x ) = 2 n + 1 . On the contrary, suppose f ( x ) does not have a zero in the field of real numbers. That is, all the zeros of f ( x ) are complex

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Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 8.4, Problem 29E

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To prove: Any polynomial of odd degree that has real coefficients must have a zero in the field of real numbers.

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Chapter 8.4, Problem 28E

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Chapter 8 Solutions

Elements Of Modern Algebra

Ch. 8.1 - True or False Label each of the following...Ch. 8.1 - Prob. 2TFE Ch. 8.1 - Prob. 3TFE Ch. 8.1 - Prob. 4TFE Ch. 8.1 - Prob. 5TFE Ch. 8.1 - Prob. 6TFE Ch. 8.1 - Prob. 7TFE Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E

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To prove: Any polynomial of odd degree that has real coefficients must have a zero in the field of real numbers.

Solution Summary: The author explains that any polynomial of odd degree that has real coefficients must have a zero in the field of real numbers.

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To prove: Any polynomial of odd degree that has real coefficients must have a zero in the field of real numbers.

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Chapter 8 Solutions