Calculating polynomials over Galois Field GF(2)

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Possible that someone could confirm that I am doing this correctly?

Answer:

The term for $x^8 + x^4 + x^3 + x + 1$ over GF(2) in binary representation is 100011011

representation of $x^8$ is 100000000

$x^8 \bmod x^8 + x^4 + x^3 + x + 1$

(100011011) mod (100000000) = 011011

representation of $x^9$ in binary is = 1000000000

so $x^9\bmod x^8 + x^4 + x^3 + x + 1$

=> (1000000000) mod (100011011) = 010101111

representation of $x^{10}$ in binary is = 10000000000

so $x^{10} \bmod x^8 + x^4 + x^3 + x + 1$

=> (10000000000) mod (100011011) = 01000011

representation of $x^{11}$ in binary is 100000000000

so $x^{11} \bmod x^8 + x^4 + x^3 + x + 1$

=> (100000000000) mod (100011011) = 010000110

representation of $x^{12}$ in binary is 1000000000000

so $x^{12} \bmod x^8 + x^4 + x^3 + x + 1$

=> (1000000000000) mod (100011011) = 0100001100

representation of $x^{13}$ in binary is 10000000000000

so $x^{13} mod x^8 + x^4 + x^3 + x + 1$

=> (10000000000000) mod (100011011) = 011111101

representation of $x^{14}$ binary is 100000000000000

so $x^{14} \bmod x^8 + x^4 + x^3 + x + 1$

=> (100000000000000) mod (100011011) = 011111101

representation of $x^{15}$ in binary is 1000000000000000

so $x^{15} mod x^8 + x^4 + x^3 + x + 1$

=> (1000000000000000) mod (100011011) = 011011111