Show that $f(x)$ is irreducible and if $a \in \mathbb{C}$ is a root of $f(x)$ then $\frac{-1}{a+1}$ is also a root of $f(x)$

Question

Problem: Let $n$ be any integer. Consider this polynomial $$f(x) = x^3 + n x^2 + (n-3)x - 1 \in \mathbb{Q}[x]$$ Show that $f(x)$ is irreducible and if $a \in \mathbb{C}$ is a root of $f(x)$ then $\frac{-1}{a+1}$ is also a root of $f(x)$.

Could we use the Eisenstein's criterion to solve?

Answer 1

For the irreduciblility, the rational root test implies that the polynomial does not have a rational root since $1,-1$ are not roots, so it is irreducible since its degree is $3$.

For the second compute $f({1\over{a+1}})={{(-1)^3}\over{(a+1)^3}}+ +n{{(-1)^2}\over{(a+1)^2}}+(n-3){{(-1)}\over{a+1}}-1=$

${{(-1)^3}\over{(a+1)^3}}(1-n(a+1)+(n-3)(a+1)^2+(a+1)^3$

We have $(1-n(a+1)+(n-3)(a+1)^2+(a+1)^3=1-n(a+1)+$

$(n-3)(a^2+2a+1)+a^3+3a^2+3a+1=$

$=a^3+na^2+(n-3)a-1=0$.

Answer 2

Substituting $x=\frac{-1}{a+1}$ in $f(x)=0$ we obtain: $$ -a^3-na^2-(n-3)a+1=0 $$