Cryptography, Cybersecurity.
addition we get x³+x² (12–1100b), and for multiplication we get x³+x+1 (11–1011b). Add=(x³+1)+(x²+1)=x³+x² Mult=(x³+1)×(x²+1)=x⁵+x³+x²+1 Now we divide x⁵+x³+x²+1 by the primitive (x⁴+x+1) and get the remainder: __x______ x^4 + x+1 | x^5+ x^3+x^2 +1 x^5 +x^2+x -------- x^3+ x +1 Thus the result of