Unformatted text preview: f has no root, so the only possibility is that f factors into two degree 2 irreducible factors. By inspection (or apply the root theorem to y 2 + 1) f = ( x 2 + 2 )( x 2 + 3 ) , and x 2 + 2 and x 2 + 3 are irreducible. (d) If f has a degree 1 factor, then it must be of the form x + a where a = ± 1, which is not the case. Therefore either f is irreducible or it is a product of two irreducible degree 2 factors. If it is the latter, then we must have x 4 + 10 x 2 + 1 = ( x 2 + ax + d )( x 2 + bx + d ) where a , b ∈ Z and d = ± 1. The coefﬁcient of x 3 shows that a + b = 0, and then by considering the coefﬁcient of x 2 , we ﬁnd that 2 da 2 = 10, which is not possible. Therefore x 4 + 10 x 2 + 1 is irreducible over Z ....
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 Fall '07
 PALinnell
 Algebra, Polynomials, Prime number, Complex number

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