Factorising (Edexcel IGCSE Maths): Revision notes

Section 1

What does 'factorising' actually mean?

Factorising is the reverse of expanding brackets. Instead of multiplying out brackets to get a longer expression, you write an expression as a product of simpler factors.

For example, expanding gives $3x(x+2) = 3x^2 + 6x$ . Factorising reverses this: $3x^2 + 6x = 3x(x+2)$ .

Always check your answer by expanding it back out - if you get the original expression, you've factorised correctly.

Key termsfactoriseexpandfactor

Exam tip

Factorising and expanding undo each other - use expanding to check your factorised answer is correct.

Section 2

How do I factorise using a common factor?

Look for the highest common factor (HCF) of every term, including any common letters, then take it outside a bracket.

Method:

Find the HCF of the numbers.
Find the lowest power of any letter common to all terms.
Write the HCF outside the bracket, and divide each term by it to get what goes inside.

Example: $8x^2 + 12x = 4x(2x+3)$ , since $4x$ is the HCF of $8x^2$ and $12x$ .

Example with two letters: $6xy^2 - 9x^2y = 3xy(2y - 3x)$ .

Key termshighest common factor (HCF)

Common mistake

Common error: taking out a factor that isn't the HIGHEST common factor, e.g. writing $2x(4x+6)$ instead of $4x(2x+3)$ for $8x^2+12x$ . Always check the bracket has no more common factors left.

Example

Factorise $15a^3b - 10a^2b^2$ : HCF is $5a^2b$ , so the answer is $5a^2b(3a - 2b)$ .

Section 3

How do I factorise a quadratic $x^2+bx+c$ ?

When the coefficient of $x^2$ is 1, find two numbers that multiply to give $c$ and add to give $b$ .

$x^2 + bx + c = (x+p)(x+q) \text{ where } p \times q = c \text{ and } p+q=b$

Example: factorise $x^2+7x+12$ . Find two numbers that multiply to 12 and add to 7: these are 3 and 4. So $x^2+7x+12=(x+3)(x+4)$ .

Watch the signs:

If $c$ is positive, $p$ and $q$ have the same sign as $b$ .
If $c$ is negative, $p$ and $q$ have opposite signs.

Example: $x^2-2x-15$ : need two numbers multiplying to $-15$ , adding to $-2$ : these are $-5$ and $3$ . So $(x-5)(x+3)$ .

Key termsquadratic expressioncoefficient

Exam tip

List the factor pairs of $c$ systematically, then check which pair adds to give $b$ - this avoids missing the correct pair.

Common mistake

Common error: getting the signs wrong, e.g. writing $(x-3)(x-4)$ for $x^2+7x+12$ . Always expand your answer to check the middle term and sign are correct.

Section 4

How do I factorise a quadratic $ax^2+bx+c$ when $a\neq1$ ?

Use the split-the-middle-term method:

Multiply $a \times c$ .
Find two numbers that multiply to give $ac$ and add to give $b$ .
Rewrite $bx$ as the sum of two terms using these numbers.
Factorise in pairs (factor by grouping).

Example: factorise $3x^2+11x+6$ .

$a \times c = 3 \times 6 = 18$ . Need two numbers multiplying to 18, adding to 11: these are 9 and 2.
Rewrite: $3x^2+9x+2x+6$ .
Group: $3x(x+3)+2(x+3)$ .
Factor out $(x+3)$ : $(x+3)(3x+2)$ .

Always check for a common factor first - it may simplify the numbers before you split the middle term.

Key termsfactor by grouping

Think of it like this

Think of splitting the middle term like unpacking a suitcase into two smaller bags that are easier to carry (factorise) separately, then noticing both bags share a handle (the common bracket).

Example

Factorise $2x^2-5x-3$ : $ac=-6$ , numbers are $-6$ and $1$ . So $2x^2-6x+x-3 = 2x(x-3)+1(x-3) = (x-3)(2x+1)$ .

Section 5

How do I use factorising to simplify algebraic fractions?

Fully factorise the numerator and denominator, then cancel any factors that appear in both.

Example: simplify $\dfrac{x^2-9}{x^2+5x+6}$ .

Factorise the numerator (difference of two squares): $x^2-9=(x-3)(x+3)$ .
Factorise the denominator: $x^2+5x+6=(x+2)(x+3)$ .
Cancel the common factor $(x+3)$ : $\dfrac{(x-3)(x+3)}{(x+2)(x+3)}=\dfrac{x-3}{x+2}$ .

Never cancel individual terms that are added or subtracted - only cancel whole factors.

Key termsdifference of two squaressimplify

Common mistake

Common error: cancelling terms instead of factors, e.g. wrongly cancelling the $x$ in $\dfrac{x+3}{x+5}$ . You can only cancel a factor that multiplies the WHOLE of the numerator and denominator.

Must Know

Factorising is the reverse of expanding brackets - always check by expanding back.
Always look for a common factor (HCF) first, before trying any other method.
For $x^2+bx+c$ : find two numbers that multiply to $c$ and add to $b$ .
For $ax^2+bx+c$ : multiply $a \times c$ , split the middle term, then factorise by grouping.
$x^2-a^2 = (x-a)(x+a)$ - the difference of two squares - appears often in fraction simplification.
Only cancel whole factors in a fraction, never individual terms.

That's the notes covered.

Carry on to the next subtopic.

What does 'factorising' actually mean?

How do I factorise using a common factor?

How do I factorise a quadratic x2+bx+cx^2+bx+cx2+bx+c?

How do I factorise a quadratic ax2+bx+cax^2+bx+cax2+bx+c when a≠1a\neq1a=1?

How do I use factorising to simplify algebraic fractions?

Must Know

How do I factorise a quadratic $x^2+bx+c$ ?

How do I factorise a quadratic $ax^2+bx+c$ when $a\neq1$ ?