What does it mean to factor over the real numbers?

Question

I am confused on the topic of factoring over real numbers. What is the difference between normally factoring and factoring over real numbers? If anyone could explain, that would be appreciated! Thanks ahead of time!

Answer 1

Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $\pm \sqrt2$.

Answer 2

Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.

Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-\sqrt2)(x+\sqrt2)$, by difference of squares. Note that $\sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).

Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!

Hopefully this demonstrates the importance of the field you are factoring over.

Answer 3

A simple example would be factoring $x^2 - 2=0$.

This polynomial has no rational roots so it can not be factored over the rationals.

But it has two real roots $\sqrt{2}$ and $-\sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - \sqrt{2})(x + \sqrt{2})$.

A more illustrative example might be $x^3 - 2x^2 - x + 2$.

By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:

$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.

But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+\sqrt 2)(x - \sqrt 2)$.

Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)

And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.