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Unformatted text preview: x,x + 1 ,x 2 + x + 1 ,x 3 + x 2 +1 ,x 3 + x +1 ,x 4 + x 3 +1 ,x 4 + x +1 ,x 4 + x 3 + x 2 + x +1 are irreducibles. 14C.47(iii) [graded out of 3 points] x 7 + x 6 + x 4 + 1 = ( x + 1)( x 2 + x + 1)( x 4 + x 3 + 1). Note that these factors are irreducibles by problem 46 above. 17A.2 [graded out of 4 points] Units: 1 , 2 ,x + 2 , 2 x + 1. Zero divisors: x, 2 x,x + 1 , 2 x + 2. 1 17A.14(iii) [graded out of 5 points] This can be done by noting that x 9 ≡ 16 x (mod x 2 + 2). So the question becomes to ﬁnd f ( x ) such that 16 xf ( x ) ≡ 1(mod x 2 + 2). Solving this congruence gives f ( x ) =1 32 x . 2...
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 Spring '08
 BILLERA
 Algebra, Remainder, Prime number, root theorem

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