Prime factorisation and exponent notation | Year 7

01 / LEARN

Understand the idea.

Keep splitting composite factors

A prime factorisation breaks a whole number greater than 1 into a product containing only primes. After splitting a factor pair, inspect both branches. A composite factor must be split again; it is not a finished leaf.

Different starting pairs give the same prime factors, apart from their order. To check your result, multiply the final prime factors back together and recover the original number. This catches factors lost while drawing a factor tree.

A power counts equal factors

In a power such as $2^{4}$ , the base is the repeated factor and the exponent counts its copies. Here there are four factors of 2, giving 16. Exponent notation groups repeated factors without changing their product.

To find an HCF from prime factorisations, keep only shared primes. For each shared prime, use the smaller number of copies. Taking more would create a product that fails to divide one of the original numbers.

\begin{aligned} 72 & = 2 \times 2 \times 2 \times 3 \times 3 \\ = 2^{3} \times 3^{2} \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Prime factorisation of 72

Write 72 as a product of prime powers.

Worked solution

1
Split 72 into useful factors.
$72 = 8 \times 9$
Both factors can be broken into familiar prime products.
2
Continue until every factor is prime.
$\begin{aligned} 8 & = 2 \times 2 \times 2 \\ 9 & = 3 \times 3 \end{aligned}$
The factors 2 and 3 are prime, so these branches are complete.
3
Group repeated factors and check.
$72 = 2^{3} \times 3^{2}, 8 \times 9 = 72$
Three factors of 2 and two factors of 3 reproduce the original number.

Answer

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

EXAMPLE 02

Use prime factors to find an HCF

Find the highest common factor of 18 and 24.

Worked solution

1
Prime-factorise both numbers.
$\begin{aligned} 18 & = 2 \times 3^{2} \\ 24 & = 2^{3} \times 3 \end{aligned}$
Prime products show exactly which factors are available in each number.
2
Keep the shared copies.
$2^{1} \times 3^{1} = 6$
Both products contain one 2 and one 3; neither contains enough copies for a larger shared prime product.
3
Verify that 6 divides each number.
$18 \div 6 = 3, 24 \div 6 = 4$
The chosen product is a common factor, and the shared-copy argument establishes it as the highest.

Answer

03 / YOUR TURN

Try it yourself.

Check your understanding

Write $5 \times 5 \times 5 \times 2$ using a power.

Show answer & reasoning

$2 \times 5^{3}$

There are three identical factors of 5, so replace them by $5^{3}$ . The remaining factor is 2, giving $2 \times 5^{3}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Write 60 as a product of prime powers.
2 marks
Write 108 as a product of prime powers.
2 marks
Write a product of five 7s using exponent notation, then evaluate the power.
2 marks
Evaluate $2^{3} \times 5^{2}$ .
2 marks
Use prime factors to find the highest common factor of 48 and 72.
3 marks
Find the whole-number value of $n$ if $2^{n} \times 3 = 96$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Write 60 as a product of prime powers.
2 marks
Split $60 = 6 \times 10$ .
Then $60 = (2 \times 3) (2 \times 5) = 2^{2} \times 3 \times 5$ .
$2^{2} \times 3 \times 5$
Write 108 as a product of prime powers.
2 marks
Split $108 = 4 \times 27$ .
Then $4 = 2^{2}$ and $27 = 3^{3}$ .
$2^{2} \times 3^{3}$
Write a product of five 7s using exponent notation, then evaluate the power.
2 marks
Five equal factors give $7^{5}$ .
Multiply $7 \times 7 \times 7 \times 7 \times 7 = 16.807$ .
$7^{5} = 16.807$
Evaluate $2^{3} \times 5^{2}$ .
2 marks
Calculate $2^{3} = 8$ and $5^{2} = 25$ .
Then $8 \times 25 = 200$ .
200
Use prime factors to find the highest common factor of 48 and 72.
3 marks
$48 = 2^{4} \times 3$ and $72 = 2^{3} \times 3^{2}$ .
Use three factors of 2 and one factor of 3.
$2^{3} \times 3 = 24$ .
24
Find the whole-number value of $n$ if $2^{n} \times 3 = 96$ .
3 marks
Divide both sides by 3 to get $2^{n} = 32$ .
$32 = 2^{5}$ , so $n = 5$ .
$n = 5$

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum

Place value and powers of ten; efficient whole-number calculations and order of operations; factors, multiples and primes; prime factorisation and exponent notation; square numbers and square roots; estimation and rounding.

Official curriculum source ↗

In this lesson

Understand the idea.

Keep splitting composite factors

A power counts equal factors

See the method in action.

Prime factorisation of 72

Split 72 into useful factors.

Continue until every factor is prime.

Group repeated factors and check.

Use prime factors to find an HCF

Prime-factorise both numbers.

Keep the shared copies.

Verify that 6 divides each number.

Try it yourself.

Check your understanding

Keep practising