# Adv Alegbra HW Solutions 115 - a 2 = 12 or a 2 = 8, each of...

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115 (vii) f ( x ) = x 4 + x 2 + x + 1. Solution. First, f ( x ) has no rational roots, by Theorem 3.90 (the only candidates are ± 1), and so we must show it is not a product of two (irreducible) quadratics. If f ( x ) factors into quadratics, then x 4 10 x 2 + 1 = ( x 2 + ax + b )( x 2 ax + c ), where a , b , c Q (the coef f cient of x in the second factor must be a because f ( x ) has no cubic term). Expanding and equating coef f cients gives b + c a 2 =− 10 , a ( c b ) = 0 , bc = 1 . Now b = 1 / c , so that ac = ab = a / c , and ac 2 = a . Hence, either a = 0or c 2 = 1. If a = 0, then we have b + c =− 10 and bc = 1 . These equations lead to a quadratic, c 2 10 c + 1, whose roots are irrational, contradicting c Q . Therefore, a ²= 0 and b = c . Since bc = 1, this gives b = c = 1o r b = c =− 1. Now c + b a 2 =− 10, so there are only two possibilities: 2 a 2 =− 10 or 2 a 2 =− 10 .
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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 f-cients 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.

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