# The Unapologetic Mathematician

## Factoring Real Polynomials

Okay, we know that we can factor any complex polynomial into linear factors because the complex numbers are algebraically closed. But we also know that real polynomials can have too few roots. Now, there are a lot of fields out there that aren’t algebraically closed, and I’m not about to go through all of them. But we use the real numbers so much because of the unique position it holds by virtue of the interplay between its topology and its algebra. So it’s useful to see what we can say about real polynomials.

We start by noting that since the real numbers sit inside the complex numbers, we can consider any real polynomial as a complex polynomial. If the polynomial has a real root $r$, then the complex polynomial has a root $r+0i$. So all the real roots still show up.

Now, we might not have as many real roots as the degree would indicate. But we are sure to have as many complex roots as the degree, which will include the real roots. Some of the roots may actually be complex numbers like $a+bi$. Luckily, one really interesting thing happens here: if $a+bi$ is a root, then so is its complex conjugate $a-bi$.

Let’s write out our polynomial $c_0+c_1X+...+c_nX^n$, where all the $c_i$ are real numbers. To say that $a+bi$ is a root means that when we substitute it for ${X}$ we get the equation

$c_0+c_1(a+bi)+...+c_n(a+bi)^n=0$

Now we can take the complex conjugate of this equation

$\overline{c_0+c_1(a+bi)+...+c_n(a+bi)^n}=\overline{0}$

But complex conjugation is a field automorphism, so it preserves both addition and multiplication

$\overline{c_0}+\overline{c_1}\left(\overline{a+bi}\right)+...+\overline{c_n}\left(\overline{a+bi}\right)^n=\overline{0}$

Now since all the $c_i$ (and ${0}$) are real, complex conjugation leaves them alone. Conjugation sends $a+bi$ to $a-bi$, and so we find

$c_0+c_1(a-bi)+...+c_n(a-bi)^n=0$

So $a-bi$ is a root as well. Thus if we have a (complex) linear factor like $\left(X-(a+bi)\right)$ we’ll also have another one like $\left(X-(a-bi)\right)$. These multiply to give

\begin{aligned}\left(X-(a+bi)\right)\left(X-(a-bi)\right)=X^2-(a-bi)X-(a+bi)X+(a+bi)(a-bi)\\=X^2-(2a)X+(a^2+b^2)\end{aligned}

which is a real polynomial again.

Now let’s start with our polynomial $p$ of degree $n$. We know that over the complex numbers it has a root $\lambda$. If this root is real, then we can write

$p=(X-\lambda)\tilde{p}$

where $\tilde{p}$ is another real polynomial which has degree $n-1$. On the other hand, if $\lambda=a+bi$ is complex then $\bar{\lambda}$ is also a root of $p$, and so we can write

$p=(X-(a+bi))(X-(a-bi))\tilde{p}=(X^2-(2a)X+(a^2+b^2))\tilde{p}$

where $\tilde{p}$ is another real polynomial which has degree $n-2$.

Either way, now we can repeat our reasoning starting with $\tilde{p}$. At each step we can pull off either a linear term or a quadratic term (which can’t be factor into two real linear terms).

Thus every real polynomial factors into the product of a bunch of linear polynomials and a bunch of irreducible quadratic polynomials, and the number of linear factors plus twice the number of quadratic factors must add up to the degree of $p$. It’s not quite so nice as the situation over the complex numbers, but it’s still pretty simple. We’ll see many situations into the future where this split between two distinct real roots and a conjugate pair of complex roots (and the border case of two equal real roots) shows up with striking qualitative effects.

August 14, 2008 - Posted by | Algebra, Polynomials, Ring theory

1. you certainly meant “factoring complex polynomials” rather than “factoring complex numbers” in the 1st line.

Comment by Dima | August 17, 2008 | Reply

2. certainly, and certainly fixed

Comment by John Armstrong | August 17, 2008 | Reply

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