A calculus of the absurd

4.3.2 Factorising

The really key idea of factorising is this very useful rule, which is that

\begin{equation} x \times (y + z) = xy + xz \end{equation}

this is known as “distributivity” (which is not really an important word to know). We can consider some examples with mathematical structures which are familiar us to help build some intuition, for example

\begin{equation} 2 \times (4 + 3) = 2 \times 7 = 14 \end{equation}

but it is also equal to

\begin{equation} 2 \times 4 + 2 \times 3 = 8 + 6 = 14. \end{equation}

Of course, this is not a proof (and proving that this is definitely true is possible, but involves too much maths to write here).

  • Example 4.3.1 This is quite useful when doing simple arithmetic, for example if we want to calculate \(12 \times 31\), we could do this by applying the distributive property and noting that \(31 = 30 + 1\), so therefore

    \begin{align} 12 \times 31 &= 12 \times (30 + 1) \\ &= 12 \times 30 + 12 \times 1 & \text {Because of distributivity} \\ &= 12 \times 30 + 12 \\ &= 360 + 12 \\ &= 372. \end{align}

Of course, we can apply this when we do algebra (i.e. to unknown variables) not just to specific numbers. One thing that most people thing is good is to “simplify” things, which makes them easier to understand. One way we can do this is to express things in terms of “irreducible” parts - i.e. things which cannot be simplified further.

  • Example 4.3.2 Let us consider the expression

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

    and try to write it in terms of irreducible components.

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 4.3.3 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}