HW9_solns - x,x + 1 ,x 2 + x + 1 ,x 3 + x 2 +1 ,x 3 + x +1...

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MATH 3360 Sketch of solutions for Assignment 9 14A.5 ( i ) Using root theorem the answer is ‘no’ in Q [ x ]. The answer is ‘no’ in Z [ x ] since it was ‘no’ in Q [ x ]. ( ii ) Dividing x 4 + x 3 + x +4 by x - 3 gives a remainder of 115, which is 5 · 23. So, m = 5 , 23 , 115 would work. 14A.10(ii) [graded out of 2 points] Roots: 0 , 2 , 12 , 20. 14A.11 One example can be R = Z /m Z , f ( x ) = x 2 . For a given n , we let m be a product of squares of n distinct prime numbers. (What are the roots?) 14B.23 A solution is r ( x ) = x 6 + x 5 + x 2 and s ( x ) = x 4 + x 3 + x 2 + x + 1. 14B.25(i) [graded out of 3 points] gcd= 1 [Note: the back of the book has a wrong answer.] 14B.28 Can be done using Euclid’s Algorithm. 14C.44 Use Fermat’s Theorem and then the Root Theorem. 14C.46 [graded out of 3 points] This is easily done using the Root Theorem to get that
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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 nd f ( x ) such that 16 xf ( x ) 1(mod x 2 + 2). Solving this con-gruence gives f ( x ) =-1 32 x . 2...
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HW9_solns - x,x + 1 ,x 2 + x + 1 ,x 3 + x 2 +1 ,x 3 + x +1...

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