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Unformatted text preview: a 2 = 12 or a 2 = 8, each of which forces a to be irrational. Therefore, there is no such factorization, and f ( x ) is irreducible over Q . (viii) f ( x ) = x 4 10 x 2 + 1. Solution. f ( x ) has no rational roots, for the only candidates are 1. Suppose that x 4 10 x 2 + 1 = ( x 2 + ax + b )( x 2 ax + c ) in Q [ x ] (we may assume the coef f cient of x in the second factor is a because f ( x ) has no cubic term). Expanding and equating coef fcients gives the following equations: c + b a 2 = 10 a ( c b ) = bc = 1 ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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