1. Find the inverse of (x5) modulo (x8+x4 +x3+ x + 1).
a) x5+ x4+ x3+x+1
b) x5+ x4+ x3
c) x5+ x4+ x3+1
d) x4+ x3+x+1
Explanation: Finding the inverse with respect to (x8+x4 +x3+ x + 1) we get x5+ x4+ x3+1 as the inverse.
2. (6x2 + x + 3)x(5x2 + 2) in Z_10 =
a) x3 + 2x + 6
b) 5x3 + 7x2 + 2x + 6
c) x3 + 7x2 + 2x + 6
d) None of the mentioned
Explanation: (6x2 + x + 3)x(5x2 + 2) in Z_10 = 5x3 + 7x2 + 2x + 6. We can find this via basic polynomial arithmetic in Z_10.
3. Is x3 + 1 reducible over GF(2)
a) Yes
b) No
c) Can’t Say
d) Insufficient Data
Explanation: Reducible: (x + 1)(x2 + x + 1).
4. Is x3 + x2 + 1 reducible over GF(2)
a) Yes
b) No
c) Can’t Say
d) Insufficient Data
Explanation: Irreducible. On factoring this polynomial, one factor is x and the other is (x + 1), which gives us the roots x = 0 or x = 1 respectively. By substitution of 0 and 1 into this polynomial, it clearly has no roots.
5. Is x4 + 1 reducible over GF(2)
a) Yes
b) No
c) Can’t Say
d) Insufficient Data
Explanation: Reducible: (x + 1)4.
6. The result of (x2 ⊗ P), and the result of (x ⊗ (x ⊗ P)) are the same, where P is a polynomial.
a) True
b) False
Explanation: The statement is true and this is the logic used behind the multiplication of polynomials on a computer. This reduces computation time.
7. The GCD of x3+ x + 1 and x2 + x + 1 over GF(2) is
a) 1
b) x + 1
c) x2
d) x2 + 1
Explanation: The GCD of x3 + x + 1 and x2 + x + 1 over GF(2) is 1.
8. The GCD of x5+x4+x3 – x2 – x + 1 and x3 + x2 + x + 1 over GF(3) is
a) 1
b) x
c) x + 1
d) x2 + 1
Explanation: The GCD of x5+x4+x3 – x2 – x + 1 and x3 + x2 + x + 1 over GF(3) is x + 1.
9. The GCD of x3 – x + 1 and x2 + 1 over GF(3) is
a) 1
b) x
c) x + 1
d) x2 + 1
Explanation: The GCD of x3 – x + 1 and x2 + 1 over GF(3) is 1.
10. Find the 8-bit word related to the polynomial x6 + x + 1
a) 01000011
b) 01000110
c) 10100110
d) 11001010
Explanation: The respective 8-bit word is 01000011.