Exam3Solutions

# Exam3Solutions - Math 302 Modern Algebra Spring 2009 Exam 3...

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Unformatted text preview: Math 302 Modern Algebra Spring 2009 Exam 3 Solutions 1. (10 points) Prove Corollary 4.30: Every polynomial f ( x ) of odd degree in R [ x ] has a root in R . Proof: Suppose f ( x ) is an arbitrary polynomial in R [ x ] with odd degree. We need to show that f ( x ) has a root in R . Suppose f ( x ) has degree n . By Corollary 4.27, we can write f ( x ) as f ( x ) = c ( x- a 1 )( x- a 2 ) · · · ( x- a n ) for some c, a 1 , a 2 , . . . , a n ∈ C . Then f ( x ) has n roots, counting mulitplicities. By Lemma 4.28, the complex, nonreal roots must come in complex conjugate pairs. Therefore the number of complex, nonreal roots must be even. Since the total number of roots is odd, then f ( x ) must have a real root. 2. (18 points) Show that a 40 ◦ angle is not constructible. Use this to show that a regular nonagon (9-sided polygon) is not constructible. To show that a 40 ◦ angle is not constructible, it su ffi ces to show that cos 40 ◦ is not constructible. If we substitute t = 40 ◦ into the trigonometric identity cos 3 t = 4 cos 3 t- 3 cos t , we obtain cos 120 ◦ = 4 cos 3 40 ◦- 3 cos 40 ◦- 1 2 = 4 cos 3 40 ◦- 3 cos 40 ◦ Then 8 cos 3 40 ◦- 6 cos 40 ◦ + 1 = 0 . This shows that cos 40 ◦ is a root of the polynomial f ( x ) = 8 x 3- 6 x + 1. By the Rational Root Test, if r/s is a rational root of f ( x ) in lowest terms, then r | 1 and s | 8. So the possible rational roots are r/s = ± 1 8 , ± 1 4 , ± 1 2 , ± 1 ....
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## This note was uploaded on 10/20/2009 for the course MATH 302 taught by Professor Edwards during the Spring '03 term at CSU Fullerton.

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Exam3Solutions - Math 302 Modern Algebra Spring 2009 Exam 3...

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