Show polynomial is irreducible

Question

Show that $x^4 + 4x^3 + 6x^2 + 2x + 1$ is irreducible over the field of rational numbers Q. Now I've tried substituting x-1 for x to show that it is irreducible, but I'm having a hard time. This is the correct way to go about it, I know, but I don't know how to execute the problem.

Answer 1

Let $P(x)=x^4+4x+6x^2+2x+1$ then $P(x)=(x+1)^4-2x=(x+1)^2-2(x+1)+2=Q(x+1)$, with $Q(y)=y^4-2y+2$. Now Eisenstein criterion applies with $p=2$.

Answer 2

By doing the substitution $x = y - 1$ as suggested, we get $y^4 - 2 y + 2$. Now apply Eisenstein's criterion.

Answer 3

Hint: Eisenstein's criterion will still work after the substitution.

A polynomial $f(x) \in \mathbb{Q}[x]$ is irreducible $\iff f(x+c)$ is irreducible for some $c \in \mathbb{Z}$.

Answer 4

Since making the substitution $x = y - 1$ requires a "lucky guess", here's a more systematic way.

This is a polynomial of degree $4$. If it factors, it'll be either as:

• The product of two degree $2$s, or
• The product of a degree $3$ and a linear polynomial.

The latter case would imply that the polynomial has a rational root, and the rational root theorem quickly shows that that isn't the case (it shows that the only possible roots are $1$ and $-1$, which can be easily ruled out by inspection).

In the former case, you can tell from the leading and constant coefficients of your polynomial that the factorization will be in the form:

$$(x^2+ax+1)(x^2+bx+1)$$

Multiplying out and comparing coefficents, you should spot a contradiction.

Answer 5

Just take the polynomial modulo $2$. By the RRT, $x^4+1$ is irreducible so we are done.