Name: Factorising Expressions (1.5.2) | Edexcel International AS Maths: Pure 1 Revision Notes 2019 Membership
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1.5.2 Factorising Expressions

What is meant by 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

How do I factorise an expression?

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. $6 x^{3} + {3 x}^{2} - 9 x = 3 x (2 x^{2} + x - 3) = 3 x (2 x + 3) (x - 1)$

Remind me how to factorise a quadratic …

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

Why does the 'ac' method work?

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

Exam Tip

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

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Edexcel International AS Maths: Pure 1

Revision Notes

1.5.2 Factorising Expressions

What is meant by factorising expressions?

How do I factorise an expression?

Remind me how to factorise a quadratic …

Why does the 'ac' method work?

Exam Tip

Author: Paul

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