# HW14sol - q ( x + 1) = ( x 2 + 2 x + 1) + ( x + 1) + 2 = x...

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MATH 3360 Applicable Algebra. HW 14 Solutions Problem 24.4 Let x = 2 + 3, then x 2 = 5 + 2 6. Subtracting 5 from both sides and squaring, gives x 4 - 10 x + 25 = ( x 2 - 5) 2 = 4 · 6 = 24. So 2 + 3 is a solution to the following minimal polynomial p ( x ) = x 4 - 10 x + 1. Problem 24.9 (i) One possible answer for a primitive root is x + 1. (ii) If you picked x + 1 for a primitive root, then the minimal poly- nomial is q ( x ) = x 2 + x + 2. You can check this, by noting that
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Unformatted text preview: q ( x + 1) = ( x 2 + 2 x + 1) + ( x + 1) + 2 = x 2 + 1 ≡ 0 mod x 2 + 1. Problem 4.1 We know that a 3-error correcting code has minimal distance of d = 3 · 2 + 1 = 7. Also, we know that 2 ∈ Z 11 has order 10. So one possible generator polynomial is g ( x ) = ( x-2)( x-2 2 )( x-2 3 )( x-2 4 )( x-2 5 )( x-2 6 ) = = ( x-2)( x-4)( x-8)( x-5)( x-10)( x-9) . 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 University (Engineering School).

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