Let $f(x) = x^3+6x^2-12x+3$ Show that f(x) is irreducible over $\mathbb Q$ . Let $\theta$ be a real root of $f(x)$, which exists due to the intermediate value theorem. $\mathbb Q(\theta)$ consists of elements of the form $a_0+a_1\theta+a_2\theta^2$. Explain how to compute the product in this field and find the product $(7+2\theta+\theta^2)(1+\theta^2)$. Find the inverse of $(7+2\theta+\theta^2)$
My attempt: for prime p =3, f(x) satisfies Eisenten's criterion, hence f(x) is irreducible over $\mathbb Q$, hence f(x) is maximal ideal in $\mathbb Q[x]$. Hence $\mathbb Q(x) \approx\mathbb Q[x]/(f(x))$
The multiplication is performed by using the distributive law and then reducing using the rule $\theta^3=-6\theta^2+12\theta-3~~(whence~~~ \theta^4=-6\theta^3+12\theta^2-3\theta)$.
$(7+2\theta+\theta^2)(1+\theta^2)= 7+7\theta^2+2\theta+2\theta^3+\theta^2+\theta^4=\theta^4+2\theta^3+8\theta^2+2\theta+7$.
replace $\theta^4~~and~~\theta^3$ above, I got $44\theta^2-49\theta+19$ as result
But I don't know how to do next. Appreciate any tips.