On multiplying (x^6+x^4+x^2+x+1) by (x^7+x+1) in GF(2^8) with irreducible polynomial (x^8 + x^4 + x^3 + x + 1) we get(a) x^7+x^6+ x^3+x^2+1(b) x^6+x^5+ x^2+x+1(c) x^7+x^6+1(d) x^7+x^6+x+1This question was addressed to me in an internship interview.Query is from Polynomial and Modular Arithmetic in section Basic Concepts in Number Theory and Finite Fields of Cryptograph & Network Security

On multiplying (x^6+x^4+x^2+x+1) by (x^7+x+1) in GF(2^8) with irreduci

1.	On multiplying (x^6+x^4+x^2+x+1) by (x^7+x+1) in GF(2^8) with irreducible polynomial (x^8 + x^4 + x^3 + x + 1) we get(a) x^7+x^6+ x^3+x^2+1(b) x^6+x^5+ x^2+x+1(c) x^7+x^6+1(d) x^7+x^6+x+1This question was addressed to me in an internship interview.Query is from Polynomial and Modular Arithmetic in section Basic Concepts in Number Theory and Finite Fields of Cryptograph & Network Security
Answer» RIGHT answer is (C) x^7+x^6+1 For explanation I WOULD say: Multiply and Obtain the modulus we GET the POLYNOMIAL product as x^7+x^6+1.

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