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In order to access this I need to be confident with:

Prime numbers

Composite numbers

Laws of indices Square numbers and square roots Cube numbers and cube rootsSurds

Factors and multiples

Multiplication

This topic is relevant for:

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.

**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).

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

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

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

\[51 = 3 × 17\]

*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.

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

E.g.

Express the number

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

So,

\[24 = 2 × 2 × 2 × 3\]

We can write this in index form:

\[24 = 2^3 × 3\]

In order to use a factor tree:

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

Get your free factor trees worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free factor trees worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEExpress the number

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

Here we write

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

The number

3**Continue until each branch ends in a prime number**

Both the numbers

Here we have highlighted the numbers

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

4**Write 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:**

\[120 = 2^3 × 3 × 5\]

The number ^{2}

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

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

**Continue until each branch ends in a prime number**

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

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

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

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

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:

\[5 × 20 = 100\]

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

Looking further down the diagram, we can see that the number

\[20 ÷ 5 = 4\]

**Continue until each branch ends in a prime number**

The only factor pair of

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

Here we are finding the product of prime factors of

\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.

Given that

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

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

Here, the question has given us a headstart as the factor pair we can use for

**Continue until each branch ends in a prime number**

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

\[72 = 2 × 2 × 2 × 3 × 3\]

In index form:

\[72 = 2^3 × 3^2\]

**Full solution:**

\[72 = 2^3 × 3^2\]

Express the number

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

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

**Continue until each branch ends in a prime number**

Keep going…

Nearly there…

One more step…

Great! We got there!

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

\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**

Let ^{2}

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

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

**Continue until each branch ends in a prime number**

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

\[21x^2 = 3 × 7 × x × x\]

**Using addition instead of multiplication**

E.g.

When creating a factor tree for say

**Assuming a number is prime**

There are several numbers which are frequently misused as a prime number, here are a few of them:

They are usually a multiple of

**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!

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.*

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

**(5 Marks)**

Show answer

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)**

Show answer

(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)**

Show answer

**(1)**

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

**(1)**

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

**(1)**

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

**(1)**

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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