$x^5 + 3x^2 - 7x - 1$ irreducible in $\mathbb{R}[x]$?

Question

I'm trying to determine if $x^5 + 3x^2 - 7x - 1$ irreducible in $\mathbb{R}[x]$

It has no obvious rational roots.

We can't apply Eisentstein's criterion as there is no $p$ that divides $-1, 3$ and $7$.

We are not dealing with $\mathbb{Z}[x]$ so I can't define homomorphisms to $\mathbb{Z}_p[x]$ and check that the factors in those rings have the same degrees.

So what else can I do to see if it is irreducible, or not?

Answer 1

Note that $f(0) = -1 < 0$, while $f(2) = 29 > 0$. Since polynomials are very differentiable, the intermediate value theorem shows that there exists an $\alpha$ between $0$ and $2$ for which $f(\alpha) = 0$. In particular, this implies that $(x - \alpha)$ is a linear factor of $f$, which is not irreducible.

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In fact, $f(-1) > 0$, so the polynomial actually has at least three real roots.

Answer 2

Remark: If $p(x)\in \Bbb R[x]$ and $\deg{(p(x))}>2$, then $p(x)$ is reducible in $\Bbb R[x]$.

Proof. Let $p(x)\in\Bbb R[x]$ be such that $\deg{(p(x))}>2$ and suppose it is irreducible over $\Bbb R[x]$. Necessarily all of its roots are complex, therefore $[\Bbb R(\alpha)\colon \Bbb R]=\deg{(p(x))}>2$, where $\alpha$ is any root of $p(x)$.
Since $\Bbb R(\alpha)\subseteq \Bbb C$ it follows that $[\Bbb C\colon \Bbb R]>2$. This contradicts the fact that $\Bbb C=\Bbb R(i)$ and $[\Bbb R(i)\colon \Bbb R]=2$.

Answer 3

Suppose $p(x)$ is a monic polynomial of degree $n \geq 1$ with real coefficients. Then, as you know, we have a factorization $$ p(x) = (x - \lambda_1) (x- \lambda_2) \cdots (x- \lambda_n) $$ where $\lambda_i \in \mathbb{C}$ for each $i$. Let bar denote complex conjugation. Since $p(x)$ has real coefficients, we know $\overline{p(x)} = p(\overline{x})$. However, $$ p( \overline{x} ) = (\overline{x} - \lambda_1) (\overline{x}- \lambda_2) \cdots (\overline{x}- \lambda_n) $$ shows us, for each $i$, that $p(\overline{\lambda_i})=0$. In other words, if a polynomial has real coefficients, then its roots come in conjugate pairs, where the real roots are considered self-conjugate. This classical result in elementary algebra is sometimes called the conjugate root theorem.

It follows that the only monic irreducibles over $\mathbb{R}$ are

• $x-a$, where $a$ is any real number
• $(x-a)^2 + b^2$, where $a$ is any real number and $b$ is a nonzero real number.

Since your polynomial is not on this list, it is reducible to a product of factors on this list. Sketching it, we see it has three roots, so it's the product of three linear factors and a quadratic irreducible.

This would be expected as background knowledge in a course studying abstract algebra, as its usually needed far before then. (For example, in linear algebra, finding if there are real eigenvalues from a minimal polynomial.) It's probably something that slipped your mind. Sometimes we get tunnel vision :)