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Unformatted text preview: Math 461 F Spring 2011 Homework 5 Solutions Drew Armstrong Problems. A.1. Let f ( x ) = a n x n + ··· + a 1 x + a ∈ R [ x ]. If n is even , with a n > 0 and a < 0, prove that f ( x ) has at least two real roots. (Hint: Intermediate value theorem.) Consider the graph of f ( x ). Since f (0) = a < 0, we see that the yintercept of the graph is negative. On the other hand, since a n > 0 and n is even, the leading term a n x n is positive for any x . For  x  large, the term a n x n will dominate, and so we have lim x →∞ f ( x ) = + ∞ and lim x → + ∞ f ( x ) = + ∞ . If lim x →∞ f ( x ) = + ∞ , there must exist some number α < 0 where f ( α ) > 0. Since f (0) = a < 0, the Intermediate Value Theorem implies that there exists some α < β < 0 such that f ( β ) = 0. Similarly, there is some value 0 < a where f ( a ) and so there exists 0 < b < a with f ( b ) = 0. We have found two real roots. A.2. Leibniz (1702) claimed that x 4 + a 4 (for a ∈ R ) cannot be factored over R . (In modern language, he claimed that x 4 + a 4 ∈ R [ x ] is irreducible .) Prove him wrong. (Hint: What are the fourth roots of a 4 ?) First we will solve the equation x 4 + a 4 , or x 4 = a 4 . Since a 4 > 0 we can write a 4 = a 4 cis( π ) in polar form. Thus the fourth roots of a 4 will have length  a  and angles ( π + 2 πk ) / 4 for k ∈ Z . In other words, 4 p a 4 = { a  cis( π/ 4) ,  a  cis(3 π/ 4) ,  a  cis(5 π/ 4) ,  a  cis(7 π/ 4) } =  a  √ 2 (1 + i ) ,  a  √ 2 ( 1 + i ) ,  a  √ 2 ( 1 i ) ,  a  √ 2 (1 i ) ....
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This note was uploaded on 01/08/2012 for the course MATH 561 taught by Professor Armstrong during the Spring '11 term at University of Miami.
 Spring '11
 Armstrong
 Algebra, Intermediate Value Theorem

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