Adv Alegbra HW Solutions 114

Adv Alegbra HW Solutions 114 - 114 (xi) If f (x ) = g (x )h...

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Unformatted text preview: 114 (xi) If f (x ) = g (x )h (x ) in Q [x ], and if f (x ) has all its coefficients in Z, then all the coefficients of g (x ) and h (x ) also lie in Z. Solution. False. (xii) For every integer c, the polynomial (x + c)2 − (x + c) − 1 is irreducible in Q [x ]. Solution. True. (xiii) For all integers n , the polynomial x 8 + 5x 3 + 5n is irreducible in Q [ x ]. Solution. False. (xiv) The polynomial x 7 + 9x 3 + (9n + 6) is irreducible in Q [x ] for every integer n . Solution. True. 3.87 Determine whether the following polynomials are irreducible in Q [x ]. (i) f (x ) = 3x 2 − 7x − 5. Solution. There are no rational roots: the candidates are ±1, ± 5, ± 1 , ± 5 . 3 3 Therefore, f (x ) is irreducible, by Proposition 3.65. (ii) f (x ) = 350x 3 − 25x 2 + 34x + 1. Solution. Absent. (iii) f (x ) = 2x 3 − x − 6. Solution. There are no rational roots: the candidates are ± 1 , ± 1, ± 3 , ± 2, ± 3, ± 6. 2 2 Therefore, f (x ) is irreducible, by Proposition 3.65. (iv) f (x ) = 8x 3 − 6x − 1. Solution. f (x ) ≡ x 3 + x − 1 mod 7. None of the elements of F7 is a root, so that f (x ) is irreducible in F7 [x ] (because a cubic is irreducible if it has no roots, by Proposition 3.65). Therefore, f (x ) is irreducible over Q. (v) f (x ) = x 3 + 6x 2 + 5x + 25. Solution. f (x ) ≡ x 3 + x + 1 mod 2. Since there are no roots over F2 , the cubic f (x ) is irreducible over Q, by Proposition 3.65. (vi) f (x ) = x 5 − 4x + 2. Solution. f (x ) is irreducible, by the Eisenstein criterion with p = 2. ...
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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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