Question

3. Factor (mod 2) all eight polynomials of the form 23 + b2x2 + b1x + bo into polynomials that are irreducible over F2, where bi E {0,1}. For example, x3+1 = (x+1)(x2+x+1), now you do the other 7. Recall, the irreducible polynomials over F2 of degree 3 or less are x, x + 1, x2 + x + 1, 2:3 + x + 1, and 2.3 + x2 +1.

Answer 1

. 3. The eight polynomial of the form x + 2x² + bix the whore bitto, 1} are comes pouding to beat, Diabol corrispording to + A +X+1 " back, ba, bo- ii) x² + x 7 x bast, biso , bond 23 nanti x + x2 23 To biso, bo 62-0, bist, bost 0, bizt, boro A ~ 23++1 viy xex vii) ** viliy x² 12=0, 12.0, boul b2=0, bero, bo=0 The impeducible factorization of the polynomials are . iz 33+ ose ants = (2+1)(x+1) · F:62 +1(8441) Olt) +1)(x+1) L = 3724+1 = niti , sine 2=0 in ] ii) x² + x + x Fala+2+1) het, which itself inreducible iv> 23+? x2(x+1) 2X* (2+1) x² + xty, which xxx SV itself inneducible (vi) Xx? x(x+2) (x+1) 82=0 in si - so 2x=0 infa viis = (x+) (a + Rtd) as xx.x