1.5.2 Factorising Expressions

Many expressions in mathematics are written as a sum of terms
- e.g. $x^{2} + 6 x - 16$ is the sum of three the terms $x^{2}$ , $6 x$ and $- 16$
Many expressions are written as a product of factors
- e.g. $(x + 8) (x - 2)$ is the product of the two (linear) factors $x + 8$ and $x - 2$
Factorising is the process of rewriting the sum of terms as a product of factors
- The other way round is expanding

This will depend on the nature of the expression you are dealing with
In all cases the first thing to consider is if there is a factor (number and/or letter) of all terms in the expression
- e.g. $2 x^{3} + 4 x^{2} - 8 x = 2 x (x^{2} + 2 x - 4)$
A quadratic expression may be able to be factorised into two linear factors
- Look out for special cases
  - No constant term: $x^{2} + 5 x = x (x + 5)$
  - Difference of two squares (no x term and constant is square): $x^{2} - 36 = (x - 6) (x + 6)$
  - Perfect squares: $x^{2} + 8 x + 16 = (x + 4) (x + 4) = {(x + 4)}^{2}$
  - ‘Hidden’ quadratics: $3^{2 x} - 12 \times 3^{x} + 27 = {(3^{x})}^{2} - 12 (3^{x}) + 27 = (3^{x} - 3) (3^{x} - 9)$
  - More than one variable: $x^{2} - y^{2} = (x - y) (x + y)$

A cubic expression (at this level) will not contain a constant term
- This means will $x$ be a factor (and there might be a number as a factor too)
- The remaining expression will be a quadratic
  - this quadratic may also be able to be factorised
  - e.g. $6 x^{3} + {3 x}^{2} - 9 x = 3 x (2 x^{2} + x - 3) = 3 x (2 x + 3) (x - 1)$

There are many shortcuts to factorise quadratic expressions, but they often only apply under certain conditions (such as when a = 1)
- the method below works for any quadratic expression
- it is most useful when the coefficient of the $x^{2}$ term is greater than 1 (and not prime)
Follow the steps:
- STEP 1 Starting with $a x^{2} + b x + c$ find the product $a c$
  - For example: for $6 x^{2} + 7 x - 3$ $a c = 6 \times - 3 = - 18$
- STEP 2 Find two numbers m & n whose product is $a c$ and sum is $b$
  - For example: $9 \times - 2 = - 18 = a c$ & $9 + (- 2) = 7 = b$
  - So $m = 9 & n = - 2$
- STEP 3 Split the $b x$ term into $m x + n x$
  - For example: $6 x^{2} - 2 x + 9 x - 3$
- STEP 4 Factorise the first two terms and the last two terms
  - For example: $2 x (3 x - 1) + 3 (3 x - 1)$
- STEP 5 Factorise once more for the final answer
  - For example: $(3 x - 1) (2 x + 3)$

If a and/or c are prime, factorising can be done “by inspection”
- For example: the only way to split (prime) 3 into factors would be 3 and 1

Suppose $a x^{2} + b x + c \equiv (p x + r) (q x + s)$
- then expanding and simplifying gives
  - $a x^{2} + b x + c \equiv p q x^{2} + p s x + q r x + r s \equiv p q x^{2} + (p s + q r) x + r s$

By comparing coefficients
- $a = p q$
- $b = p s + q r$
- $c = r s$
Let $m = p s$ and $n = q r$ then:
- $m + n = p s + q r = b$
- $m \times n = p s q r = a c$
- Therefore these are the two numbers whose product is ac and sum is b

1-5-1-ial-fig1-we-solution-fact

Do use your tried and tested shortcuts for factorising quadratics
- We’ve explained it in full above to help you understand the process rather than to learn ‘tricks’
You don't need to learn why the 'ac' method works - but we thought you might think that the algebra is cool

Edexcel International AS Maths: Pure 1