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Unformatted text preview: x 3 + x +4 = ( x3)( x 3 +4 x 2 +12 x +37)+115, so all possible m ’s are all divisors of 115: 5 , 23 , 115. You cannot use the Remainder Theorem for the second part because Z/mZ is not necessarily a ﬁeld. p243, E11(ii) , 2 , 12 , 20 p243, E12 An easy way to do this is to take Z/p 2 Z and polynomial f ( x ) = x 2 . 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell.
 Spring '08
 BILLERA
 Math, Algebra

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