GCSE Maths Number

Factors, Multiples and Primes

Factor Trees

# Factor Trees

Here we will learn about about factor trees including how to construct factor trees and use them in a variety of contexts.

There are also factor trees worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are factor trees?

Factor trees are a way of expressing the factors of a number, specifically the prime factorization of a number.

Each branch in the tree is split into factors. Once the factor at the end of the branch is a prime number, the only two factors are itself and one so the branch stops and we circle the number.

We also must remember that 1 is not a prime number and so it will not appear in any factor tree.

Factor trees can be used to:

• find the highest common factor (HCF),
• find the lowest common multiple (LCM) (sometimes called the least common multiple)
• find other numerical properties such as whether a number is square, cube or prime

We can convert different quantities to whole numbers (kilograms to grams for example) to avoid working with decimals.
A factor tree does not contain decimals as we are working with positive factors (natural numbers).

### Prime factor trees

In order to produce a prime factor tree we need to be able to recall the prime numbers between 1 and 20.

These prime numbers are:

$2, 3, 5, 7, 11, 13, 17, 19$

Let’s have a look at an example:
E.g.
Use a factor tree to write 51 as a product prime factors

We split the original number 51 into two branches by writing a pair of factors at the end of the branch,
As 3 × 17 = 51, one branch will end in a 3, the other in 17.

Both the numbers 3 and 17 are prime numbers and so we highlight the prime numbers by circling them.

Now there is a prime number at the end of each branch we have constructed a prime factor tree.
If the numbers were not primes then we would continue to split them into factors until there was a prime number at the end of each branch.

We can now write 51 as a product of it prime factors by using the numbers at the ends of the branches:

$51 = 3 × 17$

Top tip: write the prime factors in order, smallest to largest.

The factor trees of a number are not unique, but the product of prime factors is unique.
This means that a number could have multiple different factor trees that will all give the same product of prime factors.

### Writing an answer in index form

We when write a number as a product of its prime factors we should write it in index form.

E.g.
Express the number 24 as a product of prime factors

So,

$24 = 2 × 2 × 2 × 3$

We can write this in index form:

$24 = 2^3 × 3$

By using an alternative pair of factors for 24, we can see that even though the factor tree is different, the same unique prime factorisation of 24 is given.

So,

$24 = 2 × 2 × 2 × 3$

We can write this in index form:

$24 = 2^3 × 3$

## How to use a factor tree

In order to use a factor tree:

1. Write the number at the top of the factor tree and draw two branches below
2. Fill in the branches with a factor pair of the number above
3. Continue until each branch ends in a prime number
4. Write the solution as a separate line of working (in index form if required)

## Factor tree examples

### Example 1: standard problem

Express the number 120 as a product of prime factors in index form.

1. Write the number at the top of the factor tree and draw two branches below

Here we write 120 at the top of the tree with two branches below:

2Fill in the branches with a factor pair of the number above

The number 120 has many factor pairs. As the number ends in a 0, it is easy to choose 12 × 10 so we have:

3Continue until each branch ends in a prime number

Both the numbers 12 and 10 are not prime numbers so two branches are extended from each number and filled in with their factors:

Here we have highlighted the numbers 3, 5 and 2 as they are prime numbers so the branches are terminated here. However, 4 is not a prime number so we need to continue the branches for this number only:

We now have each branch ending in a prime number and so this factor tree is now complete.

4Write the solution as a separate line of working (in index form if required).

Select each prime number and express them as a product (multiply them):

$120 = 2 × 2 × 2 × 3 × 5$

Written in index form:

$120 = 2^3 × 3 × 5$

Full solution:

$120 = 2^3 × 3 × 5$

### Example 2: index from

The number 242 can be written in the form 242 = a × b2. State the values of a and b. Show your working.

\begin{aligned} &242 = 2 × 11 × 11 \\\\ &242 = 2 × 11^2 \\\\ &a = 2, \; b=11 \end{aligned}

### Example 3: unique factor tree

Complete the diagram to show the prime factor decomposition of the missing number in index form.

The number at the top of the tree is the product of the branches below. This means that the number we are expressing is equal to:

$5 × 20 = 100$

Looking further down the diagram, we can see that the number 20 is split into two factors. One of those factors is equal to 5 therefore the other factor must be:

$20 ÷ 5 = 4$

The only factor pair of 4 that does not include 1 is 2 × 2 = 4 therefore the two values in the branches are equal to 2:

Here we are finding the product of prime factors of 100, so:

\begin{aligned} 100 &= 2 × 2 × 5 × 5\\\\ 100 &= 2^2 × 5^2 \end{aligned}

Full solution:

$100 = 2^2 × 5^2$

The factor tree in this question is unique because it was already partially completed.

### Example 4: multiple steps

Given that 72 = 9 × 8, write 72 as a product of prime factors in index form.

Here, the question has given us a headstart as the factor pair we can use for 72 is 9 × 8.

$72 = 2 × 2 × 2 × 3 × 3$

In index form:

$72 = 2^3 × 3^2$

Full solution:

$72 = 2^3 × 3^2$

### Example 5: powers of 2

Express the number 64 as a power of 2.

Keep going…

Nearly there…

One more step…

Great! We got there!

\begin{aligned} &64 = 2 × 2 × 2 × 2 × 2 × 2 \\\\ &64 = 2^6 \end{aligned}

Here are two alternatives to the same factor tree:

Version 1

Version 2

### Example 6: algebraic prime factorisation

Let x be a prime number. Express 21x2 as a product of prime factors. Do not simplify your answer.

$21x^2 = 3 × 7 × x × x$

### Common misconceptions

E.g.
When creating a factor tree for say 26, a common mistake is to write the factors of 26 as 13 and 13. This is incorrect as 13 × 13 = 169 giving the prime factor decomposition of 169, not 26.

• Assuming a number is prime

There are several numbers which are frequently misused as a prime number, here are a few of them:
1, 9, 15, 21, 27

They are usually a multiple of 3 unless they are more difficult to split into factors, such as 57 and 91. (57 = 3 × 19, 91 = 7 × 13).

• Not writing the final solution

After completing the factor tree, you must write the number as a product of its factors, otherwise you have demonstrated a method but not answered the question (such as using grid multiplication and not adding up the values in the grid for your final solution).

• Incorrect simplifying of solution

Once you have reached a prime number in the factor tree, highlight it, otherwise it can get lost in the complexity of the factor tree.
Space out the diagram so you can clearly see all the factors and circle the prime factors for your solution. Then carefully check how many of each prime number exist, then write the solution using index form. The order of the product of prime factors does not matter but the numbers do!

### Practice factor tree questions

1. Express the number 140 as a product of primes. Write your answer in index form.

2 \times 5 \times 14

10 \times 14

2^{2} \times 5 \times 7

1 \times 2^{2} \times 5 \times 7

140 = 2 \times 2 \times 5 \times 7

140 = 2^2 \times 5 \times 7

2. Write 330 as a product of primes.

2 \times 5 \times 33

2 \times 3^{2} \times 11

2^{3} \times 3^{2} \times 5^{3}

2 \times 3 \times 5 \times 11

330 = 2 \times 3 \times 5 \times 11

3. Find the values of a, b, and c where 84 = a^{2} \times b \times c

a=2, \;b=3, \;c=7

a=2, \;b=1, \;c=21

a=2,\; b=3, \;c=18

a=2, \;b=5, \;c=4

84 = 2 \times 2 \times 3 \times 7

84 = 2^2 \times 3 \times 7

a=2, \;b=3, \;c=7

4. Spot the mistake in the following calculation.

196 = 7 \times 7 \times 7 \times 7

196 = 7^4

196 should be split into 2 and 98

14 should be split into 2 and 7

7 should be split into 1 and 7

There is no mistake – the solution is correct

14=7 \times2 so 14 should be split into 2 and 7 . The tree should look like this:

196 = 7 \times 2 \times 7 \times 2

196 = 2^2 \times 7^2

( 196 could be split into 2 and 98 but 14 and 14 is also correct)

5. Write the number 243 as a power of 3.

3^{4}

3^{5}

3^{6}

3^{7}

243 = 3 \times 3 \times 3 \times 3 \times 3

243 = 3^5

6. Assume a and b are prime numbers. Expand fully 35a^2b.

35 \times a \times a \times b

5 \times 5 \times a \times b

5 \times 7 \times a \times a \times b \times b

5 \times 7 \times a \times a \times b

35a^2b = 5 \times 7 \times a \times a \times b

Remember: when we write algebraic expressions, we write numbers and then letters in alphabetical order.

### Factor trees GCSE questions

1.  Using 36 = 2^{2} \times 3^{2} state the prime factor decomposition of 720 . Show all your working.

(5 Marks)

720 = 36 × 20 (could be shown in the factor tree)

(1)

(1)

(1)

720 = 2^{2} \times 3^{2} \times 2 \times 2 \times 5

(1)

720 = 2^{4} \times 3^{2} \times 5

(1)

2.  (a) Express 900 as a product of prime factors in index form.

(b) Use part (a) to show that 900 is a square number.

(5 Marks)

(a)

(factorises two numbers correctly)

(1)

(complete factorisation)

(1)

900 = 2^{2} \times 3^{2} \times 5^{2}

(1)

(b)

900=(2 \times 3 \times 5)^{2}

(1)

\begin{aligned} 900=30^{2} \\\\ \sqrt{900}=30 \end{aligned}

(1)

3.  Simplify fully \sqrt{180}

(4 Marks)

(1)

180=2^{2}\times3^{2}\times5

(1)

\sqrt{180}=2 \times 3 \times \sqrt{5}

(1)

\sqrt{180}=6\sqrt{5}

(1)

## Learning checklist

You have now learned how to:

• use the concepts and vocabulary of prime numbers, factors (or divisors), prime factorisation, including using product notation and the unique factorisation property.

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