### Calculating with letters: Computing with letters

### Factorisation

Factoring expressions

The natural number \(6\) can be written as the product of the factors \(2\) and \(3\). A similar thing can be done for algebraic expressions.

By reading the distributive properties in reverse order you get: \[\begin{aligned}ab+ac&=a(b+c)\\[0.2cm] ac+bc&=(a+b)c\\[0.2cm] ab-ac&=a(b-c)\\[0.2cm] ac-bc&=(a-b)c\end{aligned}\] Herewith you can take a term outside brackets and obtain an expression that has two or more factors.

**Examples**

\[\begin{aligned}-4a+12&=4\cdot -a+4\cdot 3\\ &=4(-a+3)\\ \\6x^3-2x^2&=2x\cdot 3x^2-2x^2\\ &=x^2(3x-1)\\ \\3pq-6pr+3p^2&=3p\cdot q-3p\cdot 2r +3p\cdot p\\ &=3p(q-2r+p) \end{aligned}\]

We discuss two additional examples.

\[\begin{aligned}a^3-a&=a(a^2-1)\\ &=a(a+1)(a-1)\\ \\ 2a^3b^3+6ab^5&= 2ab^3(a^2+3b^2)\end{aligned}\]

The first example illustrates again that there are several possibilities: the first notation is created by recognising common terms in the original expression, but the second format is not so evident. When we take factors outside brackets, we restrict ourselves in the exercises to the first kind of factorisation: only instantly recognisable factors are factored out and placed on the left-hand side.

In the second example, there is no doubt that in \(a^2+3b^2\) are no other factors to be taken outside the brackets (at least if you compute with real numbers). This entire process is called **factorisation**. The factored form is unique up to ordering of the terms in the product and multiplication by constants.