Answer:
8a + 24
Step-by-step explanation:
Use Distributive property: a(b +c) = a*b + a*c
Here a = 8 ; b= a & c = 3
8(a + 3) = 8*a + 8*3
= 8a + 24
Answer:
Step-by-step explanation:
I am sorry but please give detailed question
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)
Step-by-step explanation:
you must have made a typo here.
none of the answer options are correct for the given problem.
let me just show you what you told me, and how I can solve at least this :
2a/4 = b/4
this simply means (multiply both sides by 4) :
2a = b
so, 5b is then (multiplying both sides by 5) :
5×2a = 5b
10a = 5b
but again, "10a" is not among the offered answers.
so, I don't know if you made a mistake with the basic problem or with the answer options.