# Polynomial factorization into irreducibles over $\mathbb{Q}[x]$

I need to find irreducible factors of $f(x)=x^4+3x^3+2x^2+1$ in $\mathbb{Q}[x]$ and explicitely prove that these factors are indeed irreducible.

I believe we can't reduce $f(x)$ any further but I have to prove that this is the case.

I have already shown that there can't be any linear factors if there was such a factorization using the rational root theorem. But another possibilty is a factorization into two polynomials of degree $2$. So something like: $f(x)=(ax^2+bx+c)(mx^2+nx+p)$. I tried writing it out but since we're working with rationals I find it difficult to find or rule out such polynomials.

What do I have to do in this case? And more general: is there a way to find such second degree factors more easily in $\mathbb{Q}$?

• – idk Jan 11 '17 at 13:04

## 3 Answers

If $x^4+3x^3+2x^2+1$ were reducible in $\mathbb Z[X]$ then it would also be reducible modulo $2$.

But $x^4+x^3+1$ is irreducible over $\mathbb Z_2$ (there are only a few possible factors to check), so $x^4+3x^3+2x^2+1$ is irreducible over $\mathbb Z$.

Therefore, due to Gauß's Lemma, $x^4+3x^3+2x^2+1$ is also irreducible over $\mathbb Q$.

.Here's a wonderful result that will tell you about the irreducibility of this polynomial, but it still requires far more work:

Cohn criteria: The polynomial $$a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$$ , where $$0 \leq a_i \leq 9$$ , is irreducible if the number $$a_na_{n-1} \ldots a_0$$ is prime, in base $$10$$. This is true, for any base greater than $$2$$ as well, as long as the $$a_i$$ are smaller than the base.

Suppose we take base four, then the coefficients give the base 4 number $$10321$$ which is $$1 + (2 \times 4) + (3 \times 16) + (1 \times 4096) = 4153$$ in decimals, which you can verify is prime easily (it is still quite small, this process took me under a minute to verify). Hence, by the Cohn criteria, this works out to be irreducible. Surely there are better criteria than this one, but somehow I felt inclined to use this one.

In general, you have to find the roots through the "quartic formula", but then this is very hard to apply. Normally when this kind of a question is given to me, I look around to using Eisenstein or Cohn or Perron criteria (you can read these up, there's a huge literature on it). Either way, it will take effort to decode quartic polynomials in general.

$$f$$ has degree $$4$$ and assumes prime values for these $$9 > 2 \cdot 4$$ points and so must be irreducible: $$\begin{array}{rl} n & f(n) \\ -8 & 2689 \\ -7 & 1471 \\ -4 & 97 \\ -3 & 19 \\ 1 & 7 \\ 3 & 181 \\ 5 & 1051 \\ 6 & 2017 \\ 7 & 3529 \\ \end{array}$$

• – lhf Oct 4 '18 at 12:12