A calculus of the absurd

4.3.2 Factorising

I’m very sorry, but it’s been a while since I learnt how to factorise, and I cannot really recommend this as an introduction to the subject. If you have no idea what factorising is, please look somewhere else!

The key principle of factorising is this

  • Definition 4.3.1 An operation (which we will refer to as \(\times \)) is distributive over another operation (which we will refer to as \(+\)) if for all \(a, b, c\) in the space of mathematical objects we are working in (read: all numbers \(x\) and \(y\)) we have

    \begin{equation} a \times (b + c) \equiv a \times b + a \times c. \label {def distributivity} \end{equation}

    We use the “\(\equiv \)” symbol to indicate this is true all the time (whereas we usually use \(=\) to specify that we are interested in specific values that make the equation true, e.g. \(x = 7\) is not true all the time - in fact it is not true most of the time!)

This is obviously a rather formal definition (well, depending on who you ask). The basic idea is that starting from this property we can deduce that a bunch of other properties must also be true. We will assume that this property is true for all numbers we encounter, and proceed.

  • Definition 4.3.2 We say an expression is factorised if we have reduced it to the product of “irreducible” expressions.

This definition is useless, I put it there to make me feel better about, well I’m not actually sure what about (it’s complicated). To actually introduce the concept, let us consider some examples.

  • Example 4.3.1 Factorise the expression

    \begin{equation} (x + x^2) \end{equation}

We can apply the distributive law here. We take the identity from Equation \(\ref {def distributivity}\) and apply it by setting \(a = x\) (confusing, I know), \(b = 1\) and \(c = x\), from which we deduce that

\begin{align} (x + x^2) = x(1 + x) \end{align} A useful factorising identity

This shows up in a number of places, especially in linear algebra.

  • Example 4.3.2 Factorise

    \begin{equation} yz (z - y) + zx (x - z) + xy (y - x). \end{equation}

We start with an excellent German word, which unfortunately doesn’t seem to have a direct English translation, called an “Ansatz”. Technically the noun shouldn’t be capitalised (in English), but I have enough problems remembering to spell nouns with capital letters in German that I don’t want to create any more for myself.

The word means something like an educated guess, or a hypothesis; essentially we guess that the answer has some kind of form, and then try to use our guess to end up with the desired result.

In our case, our "guess" will be that we can factorise this into three brackets. Why three? Well, if we have two brackets, then we will get four terms (expand \((a + b)(c + d)\) by hand if you’re not convinced that this is true), and four is (we guess that the four terms won’t split apart to give the desired number) too few. Let’s expand the expression (it makes it easier, at least in my view), to obtain this

\begin{equation} yz (z - y) + zx (x - z) + xy (y - x) = yz^2 + xy^2 + x^2z - y^2z - xz^2 - x^2y. \end{equation}

Then if we want to make three brackets, we can start by writing out our expected result as

\begin{equation} (\_ + \_)(\_ + \_)(\_ + \_) \end{equation}

Then, to get the \(yz^2\) term, we can try

\begin{equation} (y + \_)(z + \_)(z + \_) \end{equation}

Then, to get the \(xy^2\) term, we can try something like

\begin{equation} (y + \_)(z - y)(z - x) \end{equation}

Why \(-y\) and \(-x\) rather than \(+y\) and \(+x\)? Because we do have some negative terms later, and we’d be stuck if everything is positive! We can continue along these lines, and to fit the \(x^2z\) term we need

\begin{equation} (y - x)(z - y)(z - x) \end{equation}

By expanding the brackets, we can check that this means

\begin{equation} yz (z - y) + zx (x - z) + xy (y - x) = yz^2 + xy^2 + x^2z - y^2z - xz^2 - x^2y = (y - x)(z - y)(z - x). \end{equation}