Factorising by Grouping

Factorise four-term expressions by splitting them into two pairs that share the same bracket.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024 · 29 real past-paper questions below

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What Grouping Is

Some expressions have four terms and nothing shared by all four. You can still factorise them.

Split them into two pairs. Factorise each pair. If both pairs give the same bracket, that bracket comes out once.

Rule

Pairs first, then the shared bracket. The answer is always two brackets multiplied.

Key idea

Check by expanding: your two brackets must give all four terms back.

Two Pairs, One Bracket

The usual 2-mark question: factorise four terms like $5 x y + 10 x + 3 y + 6$ .

Worked example

1
$(5 x y + 10 x) + (3 y + 6)$ — Split the four terms into two pairs. The first two both contain 5x. The last two both contain 3.
2
$5 x (y + 2) + (3 y + 6)$ — Take 5x out of the first pair: $5 x y \div 5 x = y$ and $10 x \div 5 x = 2$ .
3
$5 x (y + 2) + 3 (y + 2)$ — Take 3 out of the second pair: $3 y \div 3 = y$ and $6 \div 3 = 2$ . The same bracket appears twice.
4
$(5 x + 3) (y + 2)$ — The bracket is now shared, so take it out once. What stood in front of it, 5x and 3, goes in the other bracket.
5
$(5 x + 3) (y + 2) = 5 x y + 10 x + 3 y + 6 ✓$ — Check by expanding: 5x × y, 5x × 2, 3 × y and 3 × 2 give all four terms back.

Why it works: $5 x$ lots of the bracket plus $3$ lots of it is $5 x + 3$ lots of it.

Common mistake

Stopping at $5 x (y + 2) + 3 (y + 2)$ . That is still a sum. Take the bracket out to finish.

Key idea

Always write the line $5 x (y + 2) + 3 (y + 2)$ . In a 2-mark question that line earns 1 mark on its own, even if the last step slips.

When the Third Term Is Negative

The harder question has a minus in the middle, like $4 pq - 8 p - 3 q + 6$ . Take the minus out with the second pair.

Worked example

1
$(4 pq - 8 p) + (- 3 q + 6)$ — Split into two pairs. The minus belongs to the 3q, so it goes into the second pair with it.
2
$4 p (q - 2) + (- 3 q + 6)$ — First pair: take out 4p. $4 pq \div 4 p = q$ and $- 8 p \div 4 p = - 2$ .
3
$- 3 q + 6 = - 3 (q - 2)$ — Second pair: take out −3, not 3, so the bracket matches. $- 3 q \div (- 3) = q$ and $6 \div (- 3) = - 2$ .
4
$4 p (q - 2) - 3 (q - 2)$ — Put the two pairs back together. Both now have (q − 2).
5
$(4 p - 3) (q - 2)$ — Take the bracket out once. The 4p and the −3 go in the other bracket, minus sign and all.

Check the sign: $- 3 \times - 2 = + 6$ , which is the last term.

Common mistake

Writing $- 3 (q + 2)$ . That expands to $- 3 q - 6$ , not $- 3 q + 6$ . The two brackets must be exactly the same.

Swap the Order First

Sometimes the first two terms share nothing, like $ab + 12$ in $ab + 12 + 3 a + 4 b$ . Move the terms around first.

Worked example

1
$ab + 12 + 3 a + 4 b$ — The question. ab and 12 share no factor except 1, so pairing them goes nowhere.
2
$ab + 3 a + 4 b + 12$ — Reorder: put the two terms with a next to each other, then the two with b. Adding in any order gives the same total.
3
$(ab + 3 a) + (4 b + 12)$ — Now split into two pairs. Each pair shares something.
4
$a (b + 3) + (4 b + 12)$ — First pair: take out a. $ab \div a = b$ and $3 a \div a = 3$ .
5
$a (b + 3) + 4 (b + 3)$ — Second pair: take out 4. $4 b \div 4 = b$ and $12 \div 4 = 3$ . The same bracket appears twice.
6
$(a + 4) (b + 3)$ — Take the bracket out once. The a and the 4 go in the other bracket.

Key idea

Adding is allowed in any order, so swapping terms does not change the expression.

When a Pair Shares Only 1

Sometimes the second pair has nothing to take out, like $y - 4$ in $3 x y - 12 x + y - 4$ . It is still a bracket: one lot of it.

Worked example

1
$(3 x y - 12 x) + (y - 4)$ — Split into two pairs. The first pair shares 3x. The second pair shares only 1.
2
$3 x (y - 4) + (y - 4)$ — First pair: take out 3x. $3 x y \div 3 x = y$ and $- 12 x \div 3 x = - 4$ .
3
$3 x (y - 4) + 1 (y - 4)$ — The second pair already is the bracket. Write a 1 in front, because one lot of (y − 4) is just (y − 4).
4
$(3 x + 1) (y - 4)$ — Take the bracket out once. The 3x and the 1 go in the other bracket.
5
$(3 x + 1) (y - 4) = 3 x y - 12 x + y - 4 ✓$ — Check by expanding: 3x × y, 3x × (−4), 1 × y and 1 × (−4).

Common mistake

Writing $(3 x) (y - 4)$ and losing the 1. That expands to $3 x y - 12 x$ only, so two terms are missing.

When the pair starts with a minus. Take out −1 instead, so the signs inside flip.

Worked example

1
$(5 ab - 10 a) + (- b + 2)$ — From $5 ab - 10 a - b + 2$ : split into two pairs. The minus belongs to the b, so it goes into the second pair with it.
2
$5 a (b - 2) + (- b + 2)$ — First pair: take out 5a. $5 ab \div 5 a = b$ and $- 10 a \div 5 a = - 2$ .
3
$- b + 2 = - 1 (b - 2)$ — Take out −1: $- b \div (- 1) = b$ and $2 \div (- 1) = - 2$ . Now the bracket matches.
4
$5 a (b - 2) - 1 (b - 2)$ — Both pairs now have (b − 2).
5
$(5 a - 1) (b - 2)$ — Take the bracket out once. The 5a and the −1 go in the other bracket.

Take Out a Common Factor First

“Factorise completely” means nothing may be left that still factorises. If all four terms share a number, take it out first.

Worked example

1
$6 x y + 12 x + 3 y + 6$ — The question. 6, 12, 3 and 6 all divide by 3, so 3 is shared by every term.
2
$3 (2 x y + 4 x + y + 2)$ — Take the 3 out of every term: $6 x y \div 3 = 2 x y$ , $12 x \div 3 = 4 x$ , $3 y \div 3 = y$ , $6 \div 3 = 2$ .
3
$3 [2 x (y + 2) + 1 (y + 2)]$ — Now group inside the bracket. Take 2x out of the first pair. The second pair, y + 2, is one lot of the bracket.
4
$3 (2 x + 1) (y + 2)$ — Take out the shared (y + 2). Keep the 3 in front.

Common mistake

Stopping at $(6 x + 3) (y + 2)$ . It is correct, but not complete: 6x and 3 still share 3. Take it out to get $3 (2 x + 1) (y + 2)$ .

A Bracket Is Already There

Some questions give you part of the answer already, like $4 x (p + 2 q) - pq - 2 q^{2}$ . Treat the bracket like one letter and make the rest match it.

Worked example

1
$4 x (p + 2 q) - pq - 2 q^{2}$ — The question. The first part already has the bracket. Only the last two terms need work.
2
$- pq - 2 q^{2} = - q (p + 2 q)$ — Take out −q so the bracket matches: $- pq \div (- q) = p$ and $- 2 q^{2} \div (- q) = 2 q$ .
3
$4 x (p + 2 q) - q (p + 2 q)$ — Put it back. Both parts now have (p + 2q).
4
$(4 x - q) (p + 2 q)$ — Take the bracket out once. The 4x and the −q go in the other bracket.

Back-to-front brackets. (p − 5) and (5 − p) look alike but are not equal. Take out −1 to flip one of them.

Worked example

1
$3 x (p - 5) + 2 (5 - p)$ — The question. The brackets are back to front, so they do not match yet.
2
$5 - p = - (p - 5)$ — Flip it by taking out −1. Check: $- (p - 5) = - p + 5$ , which is 5 − p.
3
$3 x (p - 5) - 2 (p - 5)$ — So $+ 2 (5 - p)$ becomes $- 2 (p - 5)$ . The sign in front changes from + to −.
4
$(3 x - 2) (p - 5)$ — Take out the shared (p − 5) once.

Common mistake

Writing $(3 x + 2) (p - 5)$ . Flipping the bracket always flips the sign in front of it too.

A Squared Bracket

A squared bracket is the bracket written twice. So it shares one copy with any other term that has that bracket.

Worked example

1
$3 x (x + 4 y) + 2 (x + 4 y)^{2}$ — The question. Both terms contain (x + 4y).
2
$3 x (x + 4 y) + 2 (x + 4 y) (x + 4 y)$ — Write the square as the bracket times itself, so you can see the copy to take out.
3
$(x + 4 y) [3 x + 2 (x + 4 y)]$ — Take one (x + 4y) out of each term. What is left: 3x from the first, 2(x + 4y) from the second.
4
$(x + 4 y) (3 x + 2 x + 8 y)$ — Expand inside the square bracket: $2 (x + 4 y) = 2 x + 8 y$ .
5
$(x + 4 y) (5 x + 8 y)$ — Collect like terms: $3 x + 2 x = 5 x$ . Done.

Key idea

Write the line $(x + 4 y) [3 x + 2 (x + 4 y)]$ . It earns a mark for working even if the tidying after it goes wrong.

Which Method? Read the Question

Count the terms first, then look at the signs.

You see	You do
Four terms, nothing shared by all	Two pairs, then the shared bracket
A minus before the third term	Take out the minus with the second pair
A pair that shares nothing	Swap the order of the terms
A pair that shares only 1	Write it as 1(bracket), or −1(bracket)
All four terms share a number	Take it out first, then group
A bracket is already there	Treat it as one letter; make the rest match
(p − 5) and (5 − p)	Flip one: 5 − p = −(p − 5)
One factor in every term	Simple Factorising
Three terms with x²	Factorising Quadratic Expressions
Square minus square	Difference of Two Squares

Key idea

If the two brackets do not match, try a different pairing or check your signs. Any pairing that works gives the same answer, and the brackets can go in either order: $(a + 4) (b + 3) = (b + 3) (a + 4)$ .

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