Answer:
Step-by-step explanation:
hello,
i advice you check the question again if it is GF(
) or GF(24). i believe the question should rather be in this form;
multiplication in GF(
): Compute A(x)B(x) mod P(x) =
+
+1, where A(x)=
+1, and B(x)=
.
i will solve the above question and i believe with this you will be able to solve any related problem.
A(x)B(x)=![(x^{2} +1) (x^{3}+x+1) mod (x^{4}+x+1 ) = (x^{5} +x^{3}+x^{2} ) + (x^{3}+x+1 ) mod (x^{4} + x+1 )](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2B1%29%20%28x%5E%7B3%7D%2Bx%2B1%29%20mod%20%28x%5E%7B4%7D%2Bx%2B1%20%20%29%20%3D%20%28x%5E%7B5%7D%20%2Bx%5E%7B3%7D%2Bx%5E%7B2%7D%20%20%29%20%2B%20%28x%5E%7B3%7D%2Bx%2B1%20%20%29%20mod%20%28x%5E%7B4%7D%20%2B%20x%2B1%20%29)
= ![x^{5}+2x^{3} +x^{2} + x + 1 mod(x^{4}+x+1 )](https://tex.z-dn.net/?f=x%5E%7B5%7D%2B2x%5E%7B3%7D%20%2Bx%5E%7B2%7D%20%20%2B%20x%20%2B%201%20mod%28x%5E%7B4%7D%2Bx%2B1%20%20%29)
=![2x^{2} +1](https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B1)
please note that the division by the modulus above we used
![\frac{x^{5}+2x^{3}+x^{2} +1 }{x^{4}+x+1}= x+\frac{2x^{3} +1}{x^{4}+x+1}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B5%7D%2B2x%5E%7B3%7D%2Bx%5E%7B2%7D%20%2B1%20%20%7D%7Bx%5E%7B4%7D%2Bx%2B1%7D%3D%20x%2B%5Cfrac%7B2x%5E%7B3%7D%20%2B1%7D%7Bx%5E%7B4%7D%2Bx%2B1%7D)
Answer:
72
Step-by-step explanation:
brainliest please
2.33.......ok, 2 is ur whole number....33 is ur fraction...the last digit is in the hundredths place....so put it over 100
ur mixed number is 2 33/100...and this does not reduce