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(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.
 Fall '11
 KeithCornell

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