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Here we will learn about prime factors including how to express a number as a product of prime factors and use the product of prime factors to recognise special numbers such as square numbers and cube numbers.

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

**Prime factors **are prime numbers that are factors of another number.

E.g.

A **composite number** is the product of two or more factors. All of these types of numbers are integers (whole numbers).

In number theory, the **Fundamental Theorem of Arithmetic** states that every integer greater than one is a prime number or can be represented by a product of prime numbers.

E.g.

\[ \begin{aligned}
240 &= 2 × 2 × 2 × 2 × 3 × 5\\\\
240 &=2^4 × 3 × 5
\end{aligned}\]

This would mean that the number **240****composite number** with the** prime factors** of **2, 3 ****5**

Expressing a composite number as a product of prime factors can be utilised for a wide variety of problems such as:

- calculating the
**highest common factor**(HCF) - calculating the
**lowest common multiple**(LCM) - simplifying
**surds** - determining whether a number is a
**square number**or**cube number** **factorising**,- calculating the
**square roots**of numbers

and much more.

We can also manipulate the prime factorisation of a number to find other numbers.

To find the **prime factors of a number**, we need to continue to divide the composite number by prime numbers until we are left with just prime factors.

Looking back at the example of

Here, the **divisor **is **3****quotient **of

Any **positive integer** can be written as a **product** of its **prime factors**. This means that we can take any positive number and write it as a series of **prime numbers** being **multiplied.**

E.g.

6 is a **product** of 2 and 3, so can be written as

9 is a **product** of 3 and 3, so can be written as

We can use a process called **prime factor decomposition **using prime factor trees in order to work out the **product of prime factors**.

E.g.

Write 36 as a product of prime factors.

So,

A question may ask you to give your answer in **index form**. To do this we write the solution using **powers**, so 36 = 2^{2}×3^{2}

E.g.

Write 54 as a product of prime factors. Give your answer in index form.

So,

In index form,

54=2×3^{2}When finding prime factors it is useful to use the divisibility rules:

In order to find prime factors of a composite number:

- Divide the composite number by a suitable prime number
- Divide the answer by another suitable prime number
- Continue until the final answer is a prime number, then state the solution

We can also use **factor trees** to help us to work out the prime factors of a number.

Get your free prime factors worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free prime factors worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Prime factors** is part of our series of lessons to support revision on **factors, multiples and primes**. You may find it helpful to start with the main **factors, multiples and primes** lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

What are the prime factors of

**Divide the composite number by a suitable prime number**

As

\[18 \div 2 = 9\]

The first prime factor of

2**Divide the answer by another suitable prime number**

\[9 \div 3 = 3\]

The second prime factor is

3**Continue until the final answer is a prime number, then state the solution**

As factors are multiplied together to achieve the composite number, we can say

\[18 = 2 × 3 × 3 \]

(or in exponent form) ^{2}.

The final solution is therefore ^{2}.

Show that

**Divide the composite number by a suitable prime number**

As

\[42 \div 2 = 21\]

The first prime factor of

**Divide the answer by another suitable prime number**

\[21 \div 3 = 7.\]

The second prime factor is

**Continue until the final answer is a prime number, then state the solution**

As factors are multiplied together to achieve the composite number, we can say

\[42 = 2 × 3 × 7 \]

An exponent form does not occur for this example as there are no powers we can simplify.

The final solution is therefore

Express

**Divide the composite number by a suitable prime number**

As

\[75 \div 5 = 15\]

The first prime factor of

**Divide the answer by another suitable prime number**

\[15 \div 5 = 3.\]

The second prime factor is

**Continue until the final answer is a prime number, then state the solution**

As factors are multiplied together to achieve the composite number, we can say

\begin{aligned}
75 &= 3 × 5 × 5 \\\\
75 &= 3 × 5^2
\end{aligned}

The final solution in index form is therefore ^{2}.

Express

**Divide the composite number by a suitable prime number**

As

\[462 \div 2 = 231\]

The first prime factor of

**Divide the answer by another suitable prime number**

The sum of the digits of

\[231 \div 3 = 77.\]

The second prime factor is

**Continue until the final answer is a prime number, then state the solution**

As

\[77 \div 7 = 11 \]

The third prime factor is

As factors are multiplied together to achieve the composite number, we can say

\[462 = 2 × 3 × 7 × 11\]

The final solution in index form is therefore

Show that

**Divide the composite number by a suitable prime number**

As the sum of the digits of

\[900 \div 3 = 300\]

The first prime factor of

**Divide the answer by another suitable prime number**

The sum of the digits of

\[300 \div 3 = 100.\]

The second prime factor is

**Continue until the final answer is a prime number, then state the solution**

As

\[100 \div 2 = 50 \]

The third prime factor is

\[50 \div 2 = 25\]

The fourth prime factor is

\[25 \div 5 = 5\]

The final two prime factors are

As factors are multiplied together to achieve the composite number, we can say

\begin{aligned}
900 &= 2 × 2 × 3 × 3 × 5 × 5 \\\\
900 &= 2^{2} × 3^{2} × 5^{2}
\end{aligned}

So

What are the prime factors of

**Divide the composite number by a suitable prime number**

As

\[4095 \div 5 = 819\]

The first prime factor of

**Divide the answer by another suitable prime number**

The sum of the digits of

\[819 \div 3 = 273.\]

The second prime factor is

**Continue until the final answer is a prime number, then state the solution**

As the sum of the digits of

\[273 \div 3 = 91 \]

The third prime factor is

\[91 \div 7 = 13 \]

The fourth prime factor is

As factors are multiplied together to achieve the composite number, we can say

\[4095 = 3 × 3 × 5 × 7 × 13\]

(in exponent form)

\[4095 = 3^2 × 5 × 7 × 13.\]

The final solution in index form is therefore ^{2} × 5 × 7 × 13

**Listing factors instead of writing them as a product (multiplying)**

Given example

**Incorrect knowledge of prime numbers**

It is easy to find a factor of a composite number and not specifically a prime factor. Take example

When we look at larger composite numbers, there is a much more clear and efficient method to expressing the prime factors of a number (see the next lesson on factor trees).

1. What is 16 as a product of its prime factors?

4 \times 4

2^{4}

1, 2, 4, 8, 16

2 \times 2 \times 4

16 \div 2 = 8

8 \div 2 = 4

4 \div 2 = 2

2 is prime so we have found the prime factors.

16=2 \times 2 \times 2 \times 2=2^{4}

2. Express 63 as a product of its prime factors.

1 \times 3^{2}\times 7

2 \times 3 \times 7

3 \times 21

3^{2}\times 7

63 \div 3 =21

21 \div 3 =7

7 is prime.

63=3 \times 3 \times 7=3^{2} \times 7

3. Express the number 246 as a product of prime factors.

2 \times 3 \times 41

2^{2} \times 3^{2} \times 7

2 \times 3 \times 5 \times 7

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

246 \div 2 =123

123 \div 3 =41

41 is prime.

246=2 \times 3 \times 41

4. A number written as the product of its prime factors is 2 \times 3 \times 5^{2} . What is the number?

100

30

900

150

2 \times 3 \times 5 \times 5=150

5. Which of the following does not show 36 as a product of its prime factors?

2^{2} \times 9

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

2 \times 2 \times 3 \times 3

They all show 36 as a product of its prime factors

Although 2^{2} \times 9=36

9 is not a prime number.

6. 720 can be expressed in the form 2^a \times 3^b \times c . What are the values of a, b and c ?

a=3,\; b=2,\; c=5

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

a=4, \;b=2, \;c=7

a=2,\; b=3,\; c=5

720 \div 2 =360

360 \div 2 =180

180 \div 2=90

90 \div 2=45

45 \div 3 =15

15 \div 3=5

5 is prime.

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

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

1. Express 126 as a product of prime factors. Write your answer in index form.

**(3 marks)**

Show answer

126 \div 2=63,63 \div 3=21 \text { and } 21 \div 3=7

**(1)**

126=2 \times 3 \times 3 \times 7

**(1)**

126=2 \times 3^{2} \times 7

**(1)**

2. (a) Express the value 456 in the form 456=\mathrm{a}^{3} \times 3 \times \mathrm{b} where a and b are prime factors. State the values of a and b in your answer.

(b) Use your answer to part (a) to write the product of primes for the value of 456 × 9 .

**(5 marks)**

Show answer

(a)

456 \div 2=228,228 \div 2=114,114 \div 2=57,57 \div 3=19**(1)**

2 \times 2 \times 2 \times 3 \times 19 \text { or } 456=2^{3} \times 3 \times 19

**(1)**

a=2, b=19

**(1)**

(b)

9=3 \times 3 \text { or } 3^{2}**(1)**

456 \times 9=2^{3} \times 3 \times 19 \times 3 \times 3=2^{3} \times 3^{3} \times 19

**(1)**

3. (a) Which of the following options represents the prime factors of 28

\quad \quad 7 × 4 \quad \quad \quad \quad \quad 2^{2} \times 7 \\\\ 1, 2, 4, 7, 14, 28, \quad \quad \quad 2, 7

(b) What is the smallest number that you can multiply 28 by to get a square number?

**(4 marks)**

Show answer

(a)

2^{2} \times 7**(1)**

(b)

\text { Increase the power of } 7 \text { by } 1**(1)**

196 \text { or } 2^{2} \times 7 \times 7 \text { implied }

**(1)**

7

**(1)**

You have now learned how to:

- use the concepts and vocabulary of prime numbers, factors (or divisors), multiples, common factors, common multiples, highest common factor, lowest common multiple, prime factorisation, including using product notation and the unique factorisation property

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