How to Factorise by Grouping Step by Step

Key takeaways

  • Always check for a highest common factor (HCF) before you try grouping. Taking it out first keeps the rest of the working smaller.
  • To factorise by grouping, split the four terms into two pairs and take the HCF out of each pair. If the two brackets match, take that bracket out as a common factor.
  • If the brackets don’t match, rearrange the terms into new pairs and try again.
  • When there is a minus sign between the pairs, take the negative out of the second pair and flip every sign inside its bracket. For example, 2xy + 6y − 3x − 9 becomes 2y(x + 3) − 3(x + 3), which is (x + 3)(2y − 3).
  • If a pair has nothing in common, take out 1 or −1. A term with nothing written in front of it has a coefficient of 1.

In this video

  1. 0:00 Factorising out a group of terms
  2. 0:42 Taking out a common bracket
  3. 5:13 When to use grouping
  4. 6:15 Grouping four terms into two pairs
  5. 7:59 Writing the pairs in a different order
  6. 9:03 Rearranging the terms before grouping
  7. 10:26 A minus sign between the pairs
  8. 12:07 Taking out 1 or −1
  9. 14:10 Regrouping a harder expression
  10. 15:53 Taking out the HCF before grouping
  11. 18:23 More practice questions
  12. 20:29 A harder question with an HCF of 2

This lesson shows you how to factorise an expression with four terms by grouping the terms into pairs. It is for high school students who can already take out the highest common factor (HCF) and are ready for the next technique.

What does factorising by grouping mean?

Watch this part from 0:00

Factors are terms that multiply together to make another term. The highest common factor (HCF) is the biggest factor that terms share. Taking out the HCF is the easiest way to factorise. Grouping is the next technique.

With grouping, you take out a group of terms held in a bracket, in place of a single term. The answer usually ends up as two sets of brackets.

As a reminder of how factorising works, both terms in 4x + 9x contain x. Take out x and put what is left in a bracket.

4x + 9x = x(4 + 9)

Expanding x(4 + 9) gives 4x + 9x again.

How do you take out a common bracket?

Watch this part from 0:42

In 5(x + 4) + y(x + 4), both terms contain the bracket (x + 4). That bracket is a common factor, so you can take it out like a single term.

Open a new bracket for what is left. Ask what (x + 4) is multiplied by in each term. The answers are 5 and y.

5(x + 4) + y(x + 4) = (x + 4)(5 + y)

Writing an expression as two brackets like this is a big part of factorising and solving quadratics later on.

The brackets must be identical

You can only take out a bracket that is exactly the same in both terms. It appears once in the answer. Taking out a factor doesn’t multiply the two brackets together.

A term on its own has a coefficient of 1

In 3a2b3(4p − q) + (4p − q), the second term is just the bracket. When nothing is written in front of a term, there is really a coefficient of 1, so 1 is what is left.

3a2b3(4p − q) + (4p − q) = (4p − q)(3a2b3 + 1)

In the same way, x(y − 5) + 8(y − 5) = (y − 5)(x + 8). Carry each sign forward with its term. Multiplication can be done in either order, so (x + 8)(y − 5) is also correct.

In harder questions the brackets aren’t written for you. You have to work out the common bracket before you take it out. That is what grouping does.

When should you use factorising by grouping?

Watch this part from 5:13

Whenever you factorise, first look for an HCF you can take out of the whole expression. If there isn’t one and the expression has four terms, that is when you try grouping.

Always try the HCF first. It is the easiest way to factorise. If you group first and then still have a factor to take out at the end, you have already worked with larger numbers and more variables than you needed to.

Steps for factorising by grouping

  1. Check for an HCF of all four terms and take it out first.
  2. Split the four terms into two pairs.
  3. Take the HCF out of each pair.
  4. If the two brackets are identical, take out the common bracket and put what is left in a second bracket.
  5. If the brackets are different, rearrange the terms into new pairs and try again.

You can check an answer by expanding the brackets. You should get the original expression back.

How do you factorise four terms by grouping?

Watch this part from 6:15

Factorise xy + 5x + 4y + 20.

Two of the terms contain x and two contain y, but nothing is shared by all four terms except 1. There is no HCF, so move to grouping.

Pair the first two terms and the last two terms. The first pair has x in common, which gives x(y + 5). The second pair has 4 in common, which gives 4(y + 5).

xy + 5x + 4y + 20

= x(y + 5) + 4(y + 5)

Both brackets are (y + 5). Take it out and put x and 4 in the other bracket.

= (y + 5)(x + 4)

Does a different order change the answer?

No. The same expression can be written as 5x + 20 + xy + 4y.

5x + 20 + xy + 4y

= 5(x + 4) + y(x + 4)

= (x + 4)(5 + y)

This is the same answer. (5 + y) and (y + 5) are identical, because addition can be done in either order. This is called the commutative property. For example, 3 + 5 and 5 + 3 both give 8.

What if the pairs don’t share a common bracket?

Watch this part from 9:03

The terms are not always in a helpful order. Sometimes you need to rearrange them so that each pair shares a factor and the brackets match.

In the video’s example there is no HCF of all four terms. Rearrange the terms so that 5mn and 5n make one pair, and 2m2 and 2m make the other.

5mn + 5n + 2m2 + 2m

Take 5n out of the first pair. 5n × m = 5mn and 5n × 1 = 5n. Take 2m out of the second pair. 2m × m = 2m2 and 2m × 1 = 2m.

= 5n(m + 1) + 2m(m + 1)

Both brackets are (m + 1), so take it out.

= (5n + 2m)(m + 1)

The order of the brackets doesn’t matter, so (m + 1)(5n + 2m) is also correct.

How do you factorise by grouping with a negative?

Watch this part from 10:26

If there is a subtraction between the two pairs, take the negative out of the second pair. When you do this, every sign inside the second bracket flips.

Factorise 2xy + 6y − 3x − 9. Leave the terms in the order they are in.

Take 2y out of the first pair. 2y × x = 2xy and 2y × 3 = 6y.

2xy + 6y = 2y(x + 3)

The second pair is −3x − 9. Take out −3. Ask what −3 has to be multiplied by to give −9. The answer is +3, because a negative multiplied by a negative gives a positive. To get −3x, multiply −3 by x.

−3x − 9 = −3(x + 3)

Now put the pairs back together and take out the common bracket.

2xy + 6y − 3x − 9

= 2y(x + 3) − 3(x + 3)

= (x + 3)(2y − 3)

You might think −3 needs to be multiplied by −x to make the x term. Remember that the minus sign belongs to the term −3x. Once you have taken out −3, only x is left.

What if a pair has no common factor?

Watch this part from 12:07

Sometimes a pair has nothing in common. Take out 1, or −1 if the pair has a minus sign in front. This works as long as the other pair has something you can take out.

In 3ab + 4b2 − 3a − 4b, the first pair has b in common. The second pair, −3a − 4b, has nothing in common, so take out −1. The signs inside the bracket flip.

3ab + 4b2 − 3a − 4b

= b(3a + 4b) − 1(3a + 4b)

= (3a + 4b)(b − 1)

In 5x3 + 2x2 + 5x + 2, the first pair has x2 in common. The second pair has nothing in common, so take out 1.

5x3 + 2x2 + 5x + 2

= x2(5x + 2) + 1(5x + 2)

= (x2 + 1)(5x + 2)

If you can’t think of anything to take out of a pair, the answer is often 1.

Does the order of terms inside a bracket matter?

Watch this part from 14:10

For addition, no. In one harder question, neither pair had anything in common in the order given. Because 9 goes into 18, pair −9 with −18y and rewrite the expression as 2xy + x − 9 − 18y.

2xy + x − 9 − 18y

= x(2y + 1) − 9(1 + 2y)

(2y + 1) and (1 + 2y) are the same bracket, so take it out.

= (2y + 1)(x − 9)

The usual convention is to write the variable with the highest power first and work down to the constant. That makes (2y + 1) the neater way to write it.

Subtraction is different

(2y − 1) and (1 − 2y) are not the same. Subtraction is not commutative the way addition is, so you can’t flip the terms around.

How do you factorise by grouping when there is an HCF?

Watch this part from 15:53

Always take out the HCF first. In 6ab + 6bc − 3ac − 3c2, every term is a multiple of 3, so take out 3.

6ab + 6bc − 3ac − 3c2

= 3(2ab + 2bc − ac − c2)

Now group the four terms inside the bracket. The 3 stays in front the whole time. The first pair has 2b in common. The second pair has −c in common. A square bracket shows which part is being grouped.

= 3[2b(a + c) − c(a + c)]

Both brackets are (a + c). Take it out and drop the square bracket.

= 3(2b − c)(a + c)

There is an unwritten multiplication sign between the 3 and each bracket, so all three parts are multiplied together. To expand it, you would have to expand all of it.

What are some factorising by grouping practice questions?

Watch this part from 18:23

Try these questions from the video before you check the answers. Look for an HCF first, then group the terms in the order they are written.

  1. ab + 3a + 2b + 6. Answer: a(b + 3) + 2(b + 3) = (a + 2)(b + 3).
  2. 2x3 − 4x2 − 3x + 6. Answer: 2x2(x − 2) − 3(x − 2) = (2x2 − 3)(x − 2). When you take −3 out of −3x + 6, the bracket needs −2, because −3 × −2 = +6.
  3. 6mn − 9m + 4n − 6. Answer: 3m(2n − 3) + 2(2n − 3) = (3m + 2)(2n − 3).

Frequently asked questions

Do you always need to check for an HCF before grouping?

Yes. Take the HCF out of the whole expression first, then group what is left inside the bracket. If you group first, you may still have a factor to take out at the end after working with larger numbers.

Does it matter which order the brackets go in?

No. (x + 4)(5 + y) and (5 + y)(x + 4) are the same, and so are (5 + y) and (y + 5). Subtraction is different, so (2y − 1) and (1 − 2y) are not the same bracket.

How does grouping connect to factorising quadratics?

Writing an expression as two sets of brackets is a big part of factorising quadratics. The next lesson covers the cross method, which is a shortcut for working with quadratics.

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