# hw8 - x 3 + x +4 = ( x-3)( x 3 +4 x 2 +12 x +37)+115, so...

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MATH 3360 HOMEWORK SOLUTION 8 (Some computational problems are omitted.) p38, 2.19 Use Theorem 2.6.1. p233, E1 x p - x and 0. p233, E2 Interestingly, f ( x ) = x works. p234, E3 Suppose f = a 0 + a 1 x + ... + a n x n and g = b 0 + b 1 x + ... + b m x m and R does not have a zero divisor, then the coeﬃcient of the highest degree in fg is a n b m , which is not zero. Conversely if R has zero divisors, say ab = 0 while a,b 6 = 0, then let f = ax , g = bx , easy to see fg = 0. p243, E5 x 4 +
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Unformatted text preview: x 3 + x +4 = ( x-3)( 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.

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