4. Show that x(x2+1)1(mod(x3+x+1)). x(x2+1).1(mod(x3+x+1)). 5. Suppose that GF(23) is defined by the irreducible polynomial x3+x+1. From 4., deduce that the multiplicative inverse 1011 of 101 in GF(23) is: [Hint: The elements of GF(23) can be represented by polynomials ax2+bx+c whose coefficients a,b,c are in GF(2), which in this case are the bits 0,1 . We can represent the polynomials by the corresponding bit-strings: e.g., x2+1101 and x+1011. Then if we reduce he product of the polynomials modulo the irreducible polynomial x3+x+ 1 , this will correspond to the product: 010101. What was the product of the polynomials congruent to? The answer should follow.]