Special project: Factoring over $\mathbb{R}$

The most important result about a polynomial $p$ over the real numbers is that it will have at least deg$(p)/2$ irreducible factors. In other words, we have the following result.

Theorem 2.11.1   If $p(x)$ is a polynomial of degree $n$ in ${\mathbb{R}}[x]$ then for each complex roots $r=\alpha+i\beta$ of $p(x)$ the conjugate $\overline{r}=\alpha-i\beta$ is also a root. Moreover, $p(x)=a(x-r_1)...(x-r_k)q_{1}(x)...q_\ell(x)$, where $0\leq k \leq n$ (in which case we say that $p(x)$ splits or factors completely over ${\mathbb{R}}$), $0\leq \ell \leq n/2$ and each $q_i(x)$ is an irreducible quadratic polynomial in ${\mathbb{R}}[x]$.

This actually follows from the fundamental theorem of algebra, though we shall delay its proof until a later chapter.

Corollary 2.11.2   If $p(x)$ is an irreducible polynomial in ${\mathbb{R}}[x]$ then $p(x)$ has either degree 1 (with one real root) or degree 2 (with no real roots).

In fact, if $\alpha+i\beta$ is any complex root of a polynomial in ${\mathbb{R}}[x]$ then its complex conjugate $\alpha-i\beta$ must also be a root. Moreover, $x^2-2x\alpha+\alpha^2+\beta^2$ is an irreducible factor of $p(x)$.

To factor a polynomial $p(x)$ over the real numbers, one may use the following procedure.

• Eliminate any multiple factors by dividing $p(x)$ by $gcd(p(x),p'(x))$, as in Lemma 2.4.6.
• Find a (possible complex) root of the resulting polynomial.
• If the root is complex, say $\alpha+i\beta$, then use the division algorithm to factor $p(x)$ into the quadratic factor $x^2-2x\alpha+\alpha^2+\beta^2$ times a polynomial of lower degree. If the root is real use the division algorithm to factor $p(x)$ into a linear factor times a polynomial of lower degree.
• Repeat this process until $p(x)$ is completely factored.