Conjugate roots and real factorization

Complex roots come in pairs

Let $P$ be a polynomial with real coefficients. Since conjugation respects sums and products, we have for every complex number $z$:

$$P\left(\overline{z}\right) = \overline{P(z)}$$

Consequently, if $z$ is a root of $P$, then so is $\overline{z}$, with the same multiplicity. The non-real roots of a real polynomial therefore always appear in conjugate pairs.

Two immediate consequences

A real polynomial of odd degree has at least one real root: since the non-real roots are counted in pairs, an odd number of roots would remain, so at least one, on the real axis. We recover, through algebra, what the intermediate value theorem gave through analysis.

And the product of two conjugate factors is real:

$$\left(X - z\right)\left(X - \overline{z}\right) = X^2 - 2\,\mathrm{Re}(z)\,X + \lvert z \rvert^2$$

Its discriminant is $4\left(\mathrm{Re}(z)^2 - \lvert z \rvert^2\right) = -4\,\mathrm{Im}(z)^2 < 0$ as soon as $z$ is not real.

Factorization in $\mathbb{R}[X]$

By grouping each non-real root with its conjugate, we obtain the statement used in practice:

Every nonconstant real polynomial factors in $\mathbb{R}[X]$ into a product of factors of degree $1$ and factors of degree $2$ with strictly negative discriminant.

This result underlies partial fraction decomposition, and therefore the computation of antiderivatives of rational functions.

A complete example

Let's factor $X^4 + 1$. Its roots are the fourth roots of $-1$:

$$z_k = e^{i\left(\frac{\pi}{4} + \frac{k\pi}{2}\right)}, \qquad k \in \{0, 1, 2, 3\}$$

that is, $\dfrac{\sqrt{2}}{2}\left(\pm 1 \pm i\right)$. In $\mathbb{C}[X]$, the polynomial is therefore a product of four factors of degree $1$. Grouping the conjugates in pairs, we obtain in $\mathbb{R}[X]$:

$$X^4 + 1 = \left(X^2 - \sqrt{2}\,X + 1\right)\left(X^2 + \sqrt{2}\,X + 1\right)$$

Both factors have discriminant $2 - 4 = -2 < 0$: they are indeed irreducible over $\mathbb{R}$. This polynomial, which has no real root at all, is nevertheless not irreducible in $\mathbb{R}[X]$.

Classic pitfall

Believing that a real polynomial with no real root is irreducible over $\mathbb{R}$. This is true in degree $2$, false from degree $4$ on, as $X^4 + 1$ shows. The only irreducible polynomials in $\mathbb{R}[X]$ are those of degree $1$ and those of degree $2$ with negative discriminant — no others.