Verified answer

If it were not irreducible, it would have to be reducible into two linear terms.

However, if that were the case, what linear terms could it be? x or x+1. Those are the ONLY degree 1 polynomials over this field. But what happens if we try those:

(x)(x) = x² ≠ f(x)

(x)(x+1) = x² + x ≠ f(x)

(x+1)(x+1) = x² + 2x + 1 = x² + 1 ≠ f(x)

So it can't be that f(x) factors into any combination of those linear factors. Since they are the only possible linear factors, QED.

a million) G could be seen a collection of equivalence training. (a,b,c) = (a,b,c) + (a million,2,4), a =0,a million, b = 0,a million,2,3,c = 0,a million,2,3,4,6,7 there is exactly one component in each and every equivalence classification that has a=0. So the gang is given by way of (0,b,c) the place b, c are unfastened in Z4 and Z8 So G = Z4 x Z8 2) (a million,2,3) = (a million,2,3) - (a million,2,4) = (0,0,-a million) has order 8 in G.