Adv Alegbra HW Solutions 116

Adv Alegbra HW Solutions 116 - 116 The middle equation...

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Unformatted text preview: 116 The middle equation gives a (c − b) = 0, so that either a = 0 or b = c. In the first case, we obtain c + b = 10 cb = 1. Substituting c = b−1 , the first equation gives b2 − 10b + 1 = 0. √ But the quadratic formula gives b = 5 ± 2 6, which is irrational. On the other hand,if b = c, then bc = 1 implies b = ±1 = c. The first equation gives a 2 = −10 ± 2 < 0, and this is also impossible. We conclude that there is no factorization of f (x ) in Q [x ]. (ix) f (x ) = x 6 − 210x − 616. Solution. Eisenstein’s criterion applies, for 7 | 210 and 7 | 616, but 72 616. (x) f (x ) = 350x 3 + x 2 + 4x + 1. Solution. Reducing mod 3 to gives an irreducible cubic in F3 [x ]. 3.88 If p is a prime, prove that there are exactly 1 p3 − p monic irreducible 3 cubic polynomials in F p [x ]. Solution. Absent. 3.89 Prove that there are exactly 6 irreducible quintics in F2 [x ]. Solution. There are 32 quintics in F2 [x ], 16 of which have constant term 0; that is, have 0 as a root. Of the 16 remaining polynomials, we may discard those having an even number of nonzero terms, for 1 is a root of these; and now there are 8. If a quintic f (x ) with no roots is not irreducible, then its factors are irreducible polynomials of degrees 2 and 3; that is, f (x ) = (x 2 + x + 1)(x 3 + x + 1) = x 5 + x 4 + 1, or f (x ) = (x 2 + x + 1)(x 3 + x 2 + 1) = x 5 + x + 1. Thus, the irreducible polynomials are: x5 + x3 + x2 + x + 1 x5 + x4 + x2 + x + 1 x5 + x4 + x3 + x + 1 x5 + x4 + x3 + x2 + 1 x5 + x3 + 1 3.90 (i) x 5 + x 2 + 1. If a = ±1 is a squarefree integer, show that x n − a is irreducible in Q [x ] for every n ≥ 1. Conclude that there are irreducible polynomials in Q [x ] of every degree n ≥ 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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