Answer to Question #271308 in Abstract Algebra for N. Kokom

Question

Show that x³+ x²+x+1 is reducible over Q. Does this fact contradict the corollary to Theorem 17.4?

Answer 1

$x^3+ x^2+x+1=(x^2+1)(x+1)$

since it splits into factors, then it is reducible over Q

corollary:

for any prime p:

$P(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+x+1$

is irreducible over Q.

in our case: $x^{p-1}=x^3\implies p=4$ is not prime

so, our result does not contradict the corollary