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Year 7

1.11 Prime factorisations

Lesson

Introduction

All whole numbers except for 1 are either prime or composite, and composite numbers can always be written as a product of primes. Finding this product (called a prime factorisation) can be very useful.

A number is prime if it has exactly two factors: 1, and itself. A number is composite if it has more than two factors. Learn more about prime numbers in our investigation:  The Sieve of Eratosthenes  .

Factor trees

One of the best ways to find a prime factorisation is by using a factor tree. We start with the number we want to investigate, draw a box around it, and draw two lines coming out of it.

A box with 12 inside and two lines coming out of it.

Here is how we might start with the number 12.

A box with 12 inside with two lines coming out connecting to 4 and 3.

We then put two numbers that multiply to make12, such as 3 and 4, at the end of each of the lines.

A 12 in a box connected to 4 in a box and 3 in a circle.

Because 3 is a prime number, we circle it. Since 4 is not a prime number, we draw a box around it instead.

A 12 in a box connected to 4 in a box and 3 in a circle. And 4 is connected to 2, and 2.

We then repeat the process with 4, which is 2 \times 2.

A 12 in a box connected to 4 in a box and 3 in a circle. And 4 is connected to 2, and 2 both in circles.

And since 2 is prime, we circle both of these numbers.

This is a completed factor tree for 12, and it tells us that 12 = 12 \times 2 \times 2 \times 3. Multiplying the circled numbers at the end of each branch together always makes the original number.

A 12 in a box connected to 6 in a box and 2 in a circle. And 6 is connected to 2, and 3 both in circles.

Factor trees are not always unique - here is another factor tree for 12.

Even though the number in the box is different, the numbers at the end of the branches will always be the same for any number - they will just be in a different order.

This image shows a factor tree. Ask your teacher for more information.

Here is a factor tree for 360.

We can therefore write:

\displaystyle 360\displaystyle =\displaystyle 3 \times 5 \times 2 \times 2 \times 2 \times 3
\displaystyle 360\displaystyle =\displaystyle 2 \times 2 \times 2 \times 3 \times 3 \times 5Rewrite the factors in ascending order
\displaystyle 360\displaystyle =\displaystyle 2^{3} \times 3^{2} \times 5Use index form

Notice that the factor tree for 12 we made earlier is a smaller part of the factor tree for 360. This is because 12 is a factor of 360, and when we write 360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 we can recognize the prime factorisation of 12 inside it: 360 = 2 \times (2 \times 2 \times 3) \times 3 \times 5.

Exploration

Try the applet below to practise forming factor trees. Use the check box to know if your answer is correct or if you need a hint.

Loading interactive...

The branch of a factor tree stops where the factor is a prime number. We do not include 1 as part of our factor tree.

Examples

Example 1

A number has the following factor tree:

The image shows a factor tree. Ask the teacher for more information.

What is this number at the top of the tree?

Worked Solution
Create a strategy

Multiply the numbers to obtain the values in the factor tree.

Apply the idea
\displaystyle \text{Number}\displaystyle =\displaystyle 2 \times 2 \times 5 \times 5 \times 7 \times 11Multiply the numbers at the end of each branch
\displaystyle \text{ }\displaystyle =\displaystyle 7700Perform the multiplication

Example 2

Write 148 as a product of prime factors in expanded form.

Worked Solution
Create a strategy

Start with dividing the number with the smallest possible prime number and repeating the same process with its quotient and so on until the quotient becomes a prime number.

Apply the idea
\displaystyle 148\div 2\displaystyle =\displaystyle 74Divide by 2
\displaystyle 74 \div 2\displaystyle =\displaystyle 37Divide the quotient by 2

Since 37 is a prime number, we stop here. So we now have the expanded form: 148= 2 \times 2 \times 37

Idea summary

A number is a prime if it has exactly two factors: 1, and itself.

A number is composite if it has more than two factors.

Factor tree starts with the number that needs to be investigated and branches out to two factors. Each composite number in the factor tree branches out to two more factors until the last row of the tree are all primes.

Find all the factors

Factor trees are useful because every number we write as we make it is a factor of the original number. We do not always see every factor appear, though - for example, 9 is a factor of 360, but it does not appear in the tree above.

To find every factor of a number we need to combine the prime factors in every possible way. First, we find the prime factorisation like we did before, such as:

12 = 2 \times 2 \times 3

We then combine these factors in every possible way. Every factor of 12 can have no 2s, one 2, or two 2s in its prime factorisation. Similarly, every factor of 12 can have no 3s, or one 3. Here we draw this out in a table:

\text{No }2\text{One }2\text{Two }2\text{s}
\text{No }3122 \times 2
\text{One }332 \times 32 \times 2 \times 3

We then perform each of the multiplications to find all the factors:

\text{No }2\text{One }2\text{Two }2\text{s}
\text{No }3124
\text{One }33612

The factors of 12 are 1,\,2,\,3,\,4,\,6, and 12.

Here is how we can do it for 360 - every factor either has 5 as a factor or it doesn't, it has between zero and two 3s, and between zero and three 2s.

\text{ }\text{No }2\text{One }2\text{Two }2\text{s}\text{Three }2\text{s}
\text{No }5\text{No }3122 \times 22 \times 2 \times 2
\text{No }5\text{One }332 \times 32 \times 2 \times 32 \times 2 \times 2 \times 3
\text{No }5\text{Two }3\text{s}3 \times 32 \times 3 \times 32 \times 2 \times 3 \times 32 \times 2 \times 2 \times 3 \times 3
\text{One }5\text{No }352 \times 52 \times 2 \times 52 \times 2 \times 2 \times 5
\text{One }5\text{One }33 \times 52 \times 3 \times 52 \times 2 \times 3 \times 52 \times 2 \times 2 \times 3 \times 5
\text{One }5\text{Two }3\text{s}3 \times 3 \times 52 \times 3 \times 3 \times 52 \times 2 \times 3 \times 3 \times 52 \times 2 \times 2 \times 3 \times 3 \times 5

This table shows all the possible ways to multiply the prime factors together. We evaluate the multiplications to find all the factors:

\text{ }\text{No }2\text{One }2\text{Two }2\text{s}\text{Three }2\text{s}
\text{No }5\text{No }31248
\text{No }5\text{One }3361224
\text{No }5\text{Two }3\text{s}9183672
\text{One }5\text{No }35102040
\text{One }5\text{One }315306030
\text{One }5\text{Two }3\text{s}4590180360

The factors of 360 are 1, 2,\,3,\,4,\,5,\,6,\,8,\,9,\,10,\,12,\,15,\,18,\,20,\,24,\,30,\,36,\,40,\,45,\,60,\,72,\,90,\,120,\,180 and 360.

Examples

Example 3

Find the factors of 20.

a

Write 20 as a product of prime factors in expanded form.

Worked Solution
Create a strategy

Draw a factor tree to obtain the factors.

Apply the idea
The image shows the factor tree of 20 that has factors of 4 and 5.

Here is the beginning of a factor tree for \\ 20 = 4 \times 5

A 20 in a box connected to 4 in a box and 5 in a circle. And 4 is connected to 2, and 2 both in circles.

Continue the factor tree by finding factors of 4, and leaving the 5 which is a prime number.

4 = 2 \times 2

So:20=2\times2\times5

b

Using your answer from part (a), list all the factors of 20.

Worked Solution
Create a strategy

We can multiply all combinations of the prime factors to find all the factors.

Apply the idea
\text{No }2\text{One }2\text{Two }2\text{s}
\text{No }51\times 1=11\times 2=22\times 2=4
\text{One }51\times 5=52\times 5=102\times 2\times 5=20

So the factors of 20 are 1,\,2,\,4,\,5,\,10,\,20.

Idea summary

We can multiply all combinations of the prime factors of a number in order to find all the factors of the same number.

Outcomes

ACMNA149

Investigate index notation and represent whole numbers as products of powers of prime numbers

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