Given an irreducible polynomial p(z) F[x] we know by Corollary 29 that the field F := F[x]/(p(x)) contains F' and a root 6 := x + I of p(z). A natural question to ask is: does the field E contain all the roots of p(x)? Here we will see that sometimes the answer is \"yes\" and sometimes \"no\". (a) Let p(z) = 224+ x+1 Zy[x] and consider the field with 4 elements E := F[x]/(p(x)) = {a+b0 | a,b Zs}. We know 6 is a root of p(x). Show that E contains the other root of p(z) as well. (b) Let p(z) = a 2 Q[z] and consider the field E := Q[z]/(p(z)) = {a + b0 + c#? | a,b,c Q}. Note that the map : E Q(3/2) defined by (a + b + c#?) = a + b/2 + (/2)? is an isomorphism of the two fields. Recall the @(+/2) does not contain the other two roots of p(z) since they are non-real complex numbers while Q(+v/2) C R. Prove that does not contain any other root of p(x) other than 6. Hint: use the isomorphism