Algebra

Expanding, factorising, the laws of indices in algebra, rearranging formulae and generating sequences.

10 min read

Algebraic fluency is the single biggest time-saver on the test. You should expand and factorise on sight, rearrange any formula, and recognise the difference of two squares.

Expanding brackets: multiply every term inside by the term outside; for two binomials use FOIL. Factorising reverses this: take out the highest common factor, or split into brackets.

Worked example

Expand and simplify $(2 x - 3) (x + 5)$ .

$= 2 x^{2} + 10 x - 3 x - 15 = 2 x^{2} + 7 x - 15.$

Difference of two squares: $a^{2} - b^{2} = (a - b) (a + b)$ . Spotting it factorises expressions like $9 x^{2} - 16 = (3 x - 4) (3 x + 4)$ instantly.

Worked example

Factorise $x^{2} - 5 x - 14$ .

Two numbers multiplying to $- 14$ and adding to $- 5$ are $- 7$ and $+ 2$ :

$x^{2} - 5 x - 14 = (x - 7) (x + 2) .$

The index laws apply to algebra too. Simplify rational expressions by factorising and cancelling common factors; never cancel individual terms across a sum.

Worked example

Simplify $\frac{x ^{2} - 9}{x ^{2} + x - 6}$ .

$= \frac{( x - 3 ) ( x + 3 )}{( x + 3 ) ( x - 2 )} = \frac{x - 3}{x - 2} .$

To change the subject, do the same operation to both sides until the wanted letter is alone. If it appears twice, collect those terms on one side and factorise it out.

Worked example

Make $r$ the subject of $A = π r^{2}$ .

$r^{2} = \frac{A}{π} \Rightarrow r = \frac{A}{π} .$

Worked example

Make $x$ the subject of $y = \frac{x + 1}{x - 2}$ .

$y (x - 2) = x + 1 \Rightarrow y x - 2 y = x + 1 \Rightarrow x (y - 1) = 1 + 2 y \Rightarrow x = \frac{1 + 2 y}{y - 1} .$

A linear (arithmetic) sequence goes up by a fixed step $d$ ; its $n$ th term is $a + (n - 1) d$ , or more simply $d n + c$ where $d$ is the common difference.

Worked example

Find the $n$ th term of $7, 11, 15, 19, \dots$

Common difference $4$ , so it starts $4 n$ ; since $4 (1) = 4$ but the first term is $7$ , add $3$ : $n th term = 4 n + 3.$

A quadratic sequence has a constant *second* difference; its $n$ th term has the form $a n^{2} + bn + c$ , where $2 a$ equals the second difference.

Exam tip

To classify a sequence fast: look at first differences (constant ⇒ linear), then second differences (constant ⇒ quadratic). The second difference equals $2 a$ .

Cancelling a term that is part of a sum, e.g. wrongly "cancelling" the $x$ in $\frac{x + 3}{x}$ .
Sign slips when expanding $(x - a) (x - b)$ .
Forgetting to factorise the subject out when it appears twice while rearranging.

Algebra

Overview

Expanding & factorising

Indices in algebra & rational expressions

Rearranging formulae

Sequences

Common mistakes