Question

Question 14 5pt In the options below, there is only one polynomial that is irreducible over the field Z2. Determine that polynomial. 1 + x2 1+ x + 25 O x + x² + x3 0 1 + x + x2 + 23 01+ x + x2 + x3 + x4 0 1 + x3

Answer 1

Sely H a EZA then We know that (1) If f(x) E Zp [1] & f(a) 0 f (1) is irreducible over Zp (degree (f(n)) = 3) (2) If f(1) E Z [u] & f(a) = 0 from Sahne a GZ . Then flue) net irreducible (oreducible) over to (degree (f(1) > 1) i f(1) = x² Zg {0, 13- $(0) = 0 & f (1) ( (103 = 1 (0)? then = f (a) for rome ora e Zg f(n) = H² is reducible (net irreducible) polynomial over Z ()$Ww= 1+ ye? , Z= {0, 13 &(o) (+(012 Ito f(1) = 14012 - H - 2 - 0 Cover Z) then f(1) = 1+12 = f(1) = 0 for some 1 -a E Z is reducible thote poreducible) polynomial over 3 (18) f(1) = 1+ H+H5 Z = {0, 1} flo) = 1+0+05 toto f(1) 1+1+1 tt = 3 = 1 (over 3 ) 7

dre may Here (in) f(x) = 1 + x +45 4 vil f (1) HH +4² +43 +44 be reducible ar irreducible pelynomial Z over And Uе without know f(1) is irreducible over Zp > f() is irreduable over o & f(x) is not irreducible over o f(21) is net irreduable over Zp (*) (iii) f(x) = 1+ 2+ 45 & ffel Here pelynomial ding doan't ratirfy Irre duciblity criterion foW = 1+HAUS not irre ducible f() is not irreducible ( wing 4 ) over Zp AHH² + +t (vi) We know that f(x). is irreducible over Q Here f(1) = 14H4H²+4440 4 (5-1) = fu) in vredecible here 4 (5-1) over ZA prime & ovi) f(1) = 1+H+H² + H² +44 is reduable over ze