Let $n\in\Bbb Z$ and let $f(x)=x^3-nx^2-(n+3)x-1$. Show that $f(x)$ is irreducible in $\Bbb Q[x]$.

Question

Let $n\in Z$ and let $f(x)=x^3-nx^2-(n+3)x-1$.

1. Show that $f(x)$ is irreducible in $\Bbb Q[x]$.
2. Show that if $r$ is a root of $f(x)$ then $-1/(1+r)$ is also a root of $f(x)$.
3. Compute the smallest field extension of $\Bbb Q$ where $f(x)$ factors completely into linear factors.

For the first part, I used Rational Root Theorem to show that $f(1)=-2n-3\ne 0$ since $n\in Z$ and $f(-1)=1\ne 0$. And any $a,b\in Z$ such that $a \nmid -1$ or $b \nmid 1$ implies $f(a/b) \ne 0$ hence $f(x)$ has no rational root and since its degree is 3, it is irreducible in $Q[x]$. Please let me know if that's correct.

For the second part, I evaluated $g(x)$ by long division where $f(x)=g(x)(x-r)$ and since if we assume $-1/(1+r)$ is a root, then $f(-1/(1+r))=0$ implying $g(-1/(1+r))=0$. But I couldn't get $g(x)=0$ when evaluating at $-1/(1+r)$. So I couldn't solve part 2).

And similarly for 3), would smallest field extension of $Q$ be Q(r,-1/(1+r),R)? where R is the third root? I couldn't get any further either. Please let me know.

Answer 1

The polynomial $f$ has three distinct real roots: $r$, $-\frac{1}{1+r}$, and $-\frac{1+r}{r}$ (why?). It follows that its splitting field is $\mathbb Q(r)$.

For the second question: show that $f(-\frac{1}{1+r})=0$ using that $f(r)=0$. It is only a matter of calculation!