Question

Show that p(x) = x
3
+ 9x + 6 is irreducible in Q[x]. Let
θ be a root of p(x). Find the inverse of 1 + θ in Q[x].

Answer 1

equation should have been: x^3 + 9x + 6

Answer 2

Ah, i was over-complicating things a lot.
Use the rational roots theorem to show its irreducible over the rationals

Answer 3

if there was a rational root, it would have to be a divisor of 6. Since 1, 2, 3, and 6 arent roots, there are no rational roots, and p(x) is irreducible over the rationals