1.02fPureCore

Factorising Expressions

Q: What if a quadratic will not factorise?

Many do not have rational factors. Check the discriminant b^2-4ac: if it is not a perfect square, there is no factorisation with whole numbers and you should use the quadratic formula or complete the square instead.

Q: Does the order of the brackets matter?

No. (x+2 ) (x+3 ) and (x+3 ) (x+2 ) are the same factorisation and both score.

Factorising runs expansion backwards: it takes a sum and writes it as a product. That matters because a product tells you something a sum cannot — when it equals zero. Every root you find, every curve you sketch, every fraction you cancel starts by getting something into factorised form.

30 min Polynomials

The big picture

The reason factorising is worth real effort is the $zero product property$ : if $A B = 0$ then $A = 0$ or $B = 0$ . No equivalent statement exists for sums. Knowing that $x^{2} + x - 6$ equals zero somewhere is not useful; knowing it equals $(x + 3) (x - 2)$ tells you immediately that the roots are $- 3$ and $2$ . The same move underlies simplifying algebraic fractions, finding vertical asymptotes, integrating by partial fractions and solving inequalities. Factorised form is where the information lives.

What you'll be able to do

Take out the highest common factor, including negative and algebraic ones
Factorise a quadratic with unit and non-unit leading coefficient
Recognise and apply the difference of two squares
Choose the right method by inspecting the number of terms and their structure

Always look for a common factor first

Before any other method, check whether every term shares a factor and take out the largest one. This is not optional tidying — it usually turns a hard factorisation into an easy one. $2 x^{2} + 10 x + 12$ looks awkward until the 2 comes out, leaving $2 (x^{2} + 5 x + 6)$ and then $2 (x + 2) (x + 3)$ .

The common factor can involve variables too: in $6 x^{3} - 9 x^{2}$ , both terms share $3 x^{2}$ , giving $3 x^{2} (2 x - 3)$ . Take the highest power of $x$ present in every term.

1Numbers: the highest common factor of 12 and 18 is 6.

2Letters: both terms contain

x^{2}

and

y

3So take out

6 x^{2} y

and see what remains in each term.

Answer

6 x^{2} y (2 x - 3 y)

Tip — "Factorise fully" means there is nothing left to extract. If your bracket still has a common factor inside it, you have not finished.

Quadratics with leading coefficient 1

For $x^{2} + b x + c$ , you need two numbers that $multiply to c$ and $add to b$ . Those two numbers become the constants in the brackets.

The signs narrow the search fast. If $c$ is positive, both numbers share a sign — the sign of $b$ . If $c$ is negative, the numbers have opposite signs and the larger in size carries the sign of $b$ . Checking signs before hunting for factors removes most of the candidates.

c = + 12

is positive and

b = - 7

is negative, so both numbers are negative.

2Negative pairs multiplying to 12:

(- 1, - 12)

(- 2, - 6)

(- 3, - 4)

3The pair adding to

- 7

- 3

and

- 4

Answer

(x - 3) (x - 4)

Why this works

Expanding $(x + p) (x + q)$ gives $x^{2} + (p + q) x + pq$ — so the "multiply to $c$ , add to $b$ " rule is not a trick, it is that expansion read backwards.

Quadratics with a leading coefficient

When $a \neq = 1$ , the reliable method is $splitting the middle term$ . Find two numbers that multiply to $a c$ and add to $b$ , use them to split $b x$ into two terms, then factorise the resulting four terms in pairs.

The method never requires guessing bracket arrangements, which is what makes it dependable under exam pressure — and it works unchanged however large the coefficients get.

a c = 6 \times 3 = 18

; find two numbers multiplying to 18 and adding to 11: that is 9 and 2.

2Split the middle:

6 x^{2} + 9 x + 2 x + 3

3Pair them:

3 x (2 x + 3) + 1 (2 x + 3)

4The bracket

(2 x + 3)

is now common.

Answer

(2 x + 3) (3 x + 1)

Tip — If the two brackets after pairing are not identical, the split was wrong or a sign slipped. Do not force it — go back and re-check the pair of numbers.

Choosing a method by looking at the expression

Most factorising errors are method-choice errors. Count the terms and check the structure first.

$Two terms$ : look for a common factor, then for a difference of two squares — but only if both terms are perfect squares and separated by a minus. A sum of two squares does not factorise over the reals.

$Three terms$ : it is a quadratic, so common factor first, then either the multiply-and-add method or the split.

$Four terms$ : try grouping in pairs, which is the same manoeuvre that finishes the split method.

a^{2} - b^{2} = (a + b) (a - b)

Difference of two squares — minus sign essential.

a^{2} + b^{2} does not factorise

Not over the real numbers.

1Common factor

2 x

2 x (4 x^{2} - 9)

2The bracket is a difference of two squares:

(2 x)^{2} - 3^{2}

Answer

2 x (2 x - 3) (2 x + 3)

Why this works

The first example is the reason "factorise fully" is worded that way. Stopping at $2 x (4 x^{2} - 9)$ is a correct factorisation and an incomplete answer — the difference of two squares inside is still waiting.

Think like an examiner

"Factorise fully" or "factorise completely" signals that more than one step is needed — usually a common factor followed by something else.
Check your factorisation by expanding mentally. It costs seconds and catches sign errors outright.
When a factorisation feeds a "solve" instruction, remember the roots come from setting each bracket to zero — and that $2 x (\dots)$ contributes the root $x = 0$ .
For a non-unit leading coefficient, show the split of the middle term. It is a credited method step even if the pairing then goes wrong.

Common misconceptions

Factorising

x^{2} + 16

(x + 4) (x - 4)

That expands to

x^{2} - 16

. A

sum

of two squares does not factorise over the reals.

Stopping at

3 (x^{2} - 4)

The bracket is a difference of two squares:

3 (x - 2) (x + 2)

For

2 x^{2} + 7 x + 3

, using numbers that multiply to 3 rather than to

a c = 6

With a leading coefficient the target product is

a c

, not

c

. Here 6 and 1 multiply to 6 and add to 7.

Cancelling

x

from

\frac{x ^{2} + x}{x}

term by term to get

x + x

Factorise first:

\frac{x ( x + 1 )}{x} = x + 1

. Cancelling only works across a factor of the whole numerator.

Factorising toolkit

ab + a c = a (b + c)

Common factor first, always.

x^{2} + (p + q) x + pq = (x + p) (x + q)

Multiply to

c

, add to

b

a^{2} - b^{2} = (a + b) (a - b)

Difference of two squares.

A B = 0 \Rightarrow A = 0 or B = 0

Why factorised form is worth having.

Stretch yourself

Factorise $x^{4} - 13 x^{2} + 36$ completely.

Hint — It is a quadratic in $x^{2}$ . Factorise it as one, then look hard at what you are left with.

Questions students ask

What if a quadratic will not factorise?

Many do not have rational factors. Check the discriminant

b^{2} - 4 a c

: if it is not a perfect square, there is no factorisation with whole numbers and you should use the quadratic formula or complete the square instead.

Is the split-the-middle-term method necessary, or can I just guess brackets?

Inspection is quicker when it works and fine for small coefficients. The split is worth learning because it always works and never leaves you searching — which matters when the numbers are unfriendly and the clock is running.

Does the order of the brackets matter?

No.

(x + 2) (x + 3)

and

(x + 3) (x + 2)

are the same factorisation and both score.

Why does factorising help solve equations?

Because a product is zero only when one of its factors is zero. That turns one hard equation into two easy ones, and nothing similar is available for a sum.

Key takeaways

Take out the highest common factor before anything else, variables included.
For $x^{2} + b x + c$ , find two numbers multiplying to $c$ and adding to $b$ .
With a leading coefficient, split the middle term using numbers multiplying to $a c$ .
Two terms suggests a common factor or a difference of two squares; four suggests grouping.
"Factorise fully" means keep going until nothing further comes out.

How this fits the course

Build on

Expanding Brackets

The Factor Theorem

Leads to

The Factor Theorem

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Real past-paper questions on Factorising Expressions, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Factorising Expressions

Always look for a common factor first

Quadratics with leading coefficient 1

Quadratics with a leading coefficient

Choosing a method by looking at the expression

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Factorising Expressions

Always look for a common factor first

Quadratics with leading coefficient 1

Quadratics with a leading coefficient

Choosing a method by looking at the expression

Practise this topic

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