On multiplying (x^5 + x^2 + x) by (x^7 + x^4 + x^3 + x^2 + x) in GF(28) with irreducible polynomial (x^8 + x^4 + x^3 + x + 1) we get(a) x^12+x^7+x^2(b) x^5+x^3+x^3(c) x^5+x^3+x^2+x(d) x^5+x^3+x^2+x+1I had been asked this question in examination.I'm obligated to ask this question of Polynomial and Modular Arithmetic in division Basic Concepts in Number Theory and Finite Fields of Cryptograph & Network Security

On multiplying (x^5 + x^2 + x) by (x^7 + x^4 + x^3 + x^2 + x) in GF(28

1.	On multiplying (x^5 + x^2 + x) by (x^7 + x^4 + x^3 + x^2 + x) in GF(28) with irreducible polynomial (x^8 + x^4 + x^3 + x + 1) we get(a) x^12+x^7+x^2(b) x^5+x^3+x^3(c) x^5+x^3+x^2+x(d) x^5+x^3+x^2+x+1I had been asked this question in examination.I'm obligated to ask this question of Polynomial and Modular Arithmetic in division Basic Concepts in Number Theory and Finite Fields of Cryptograph & Network Security
Answer» The correct ANSWER is (d) x^5+x^3+x^2+x+1 Explanation: MULTIPLICATION gives us (x^12 + x^7 + x^2) MOD (x^8 + x^4 + x^3 + x + 1). REDUCING this via modular division gives us, (x^5+x^3+x^2+x+1)

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