# Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension

## Problem 334

Prove that the polynomial

\[f(x)=x^3+9x+6\]
is irreducible over the field of rational numbers $\Q$.

Let $\theta$ be a root of $f(x)$.

Then find the inverse of $1+\theta$ in the field $\Q(\theta)$.

## Proof.

Note that $f(x)$ is a monic polynomial and the prime number $3$ divides all non-leading coefficients of $f(x)$. Also the constant term $6$ of $f(x)$ is not divisible by $3^2$. Hence by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible over $\Q$.

We divide the polynomial $f(x)$ by $x+1$ and obtain

\[x^3+9x+6=(x+1)(x^2-x+10)-4\]
by long division.

Then it follows that in the field $\Q(\theta) \cong \Q[x]/(f(x))$ (note that $f(x)$ is the minimal polynomial of $\theta$), we have

\[0=(\theta+1)(\theta^2-\theta+10)-4,\]
and hence this yields that we have the inverse

\[(1+\theta)^{-1}=\frac{1}{4}(\theta^2-\theta+10).\]

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