How to Find Prime Factors Using Factor Trees
Prime factorisation breaks a composite number into a product of prime numbers. In GCSE and IGCSE Maths, it is a foundational skill used when finding the highest common factor (HCF), lowest common multiple (LCM), simplifying fractions, and working with indices. This page explains how to use factor trees to write numbers as products of primes, with clear examples and practice to build confidence.
How to Prime Factorise a Number: Step by Step
Prime factorisation means breaking a whole number into the product of the primes that multiply together to make it. The factor-tree method is the most common approach.
- Start with the number. Write it at the top of your tree.
- Split it into any factor pair. For example, \( 90 = 2 \times 45 \) or \( 90 = 6 \times 15 \). Any valid pair works; smaller primes first usually gives a tidier tree.
- Keep splitting composite numbers. Circle any primes (they are finished), and continue splitting anything that is still composite.
- Write the final answer. Multiply all the circled primes together. Use index notation to tidy repeated primes, for example \( 156 = 2^2 \times 3 \times 13 \).
Quick example: for \(126\), one route is \( 126 = 2 \times 63 \), then \( 63 = 9 \times 7 \) and \( 9 = 3 \times 3 \). So \( 126 = 2 \times 3^2 \times 7 \).
Here is a completed tree for \(392\), and a step-by-step animation for \(156\).
One route is \(392 = 8 \times 49\), then split 8 and 49 until only primes remain.
Press Play to see each step appear. Route: \(156 = 2 \times 78\), then \(78 = 2 \times 39\), then \(39 = 3 \times 13\).
Green circle = prime, blue box = composite, orange lines = branches.
Worked Examples
💡 Example 1: When a Large Prime Appears
Write \( 110 \) as a product of its prime factors.
\( 110 = 2 \times 55 \)
What's happening?
\(110\) is even, so 2 is a factor. Divide: \( 110 \div 2 = 55 \). Circle the 2.
\( 55 = 5 \times 11 \)
What's happening?
\(55\) ends in 5, so 5 divides it. \( 55 \div 5 = 11 \), and 11 is prime. Circle both.
\( 110 = 2 \times 5 \times 11 \)
What's happening?
All factors are prime. Write the answer as a product of those primes.
💡 Example 2: Using Index Notation
Write \( 396 \) as a product of its prime factors in index form.
\( 396 = 2 \times 198 \)
\( 198 = 2 \times 99 \)
What's happening?
Divide by 2 repeatedly while the number is even. \( 396 \div 2 = 198 \) and \( 198 \div 2 = 99 \).
\( 99 = 3 \times 33 \)
\( 33 = 3 \times 11 \)
What's happening?
\(99\) is not even. Try 3 next: \( 99 \div 3 = 33 \), \( 33 \div 3 = 11 \). Both 3 and 11 are prime.
\( 396 = 2 \times 2 \times 3 \times 3 \times 11 \)
\( = 2^2 \times 3^2 \times 11 \)
What's happening?
Collect repeated primes into index form for a neat final answer.
Divisibility Quick Checks
Type any whole number from 2 to 999,999 and press Build. Primes are circles, composites are boxes.
Green circle = prime, blue box = composite, orange lines = branches.
🔑 Key Points
- Every whole number greater than 1 has a unique prime factorisation (ignoring order). This is the Fundamental Theorem of Arithmetic.
- Any valid factor pair works at each step. Different trees can give the same final answer.
- Keep splitting until every leaf is prime. A leaf is finished only when it cannot be divided further.
- Collect repeated primes into index form for your final answer: \( 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \).
- Useful for finding the HCF, LCM, simplifying surds and working with fractions.
⚠️ Common Pitfalls
- Stopping too early and leaving a composite number in the answer. \(15 \times 4\) is a factor pair but \(15\) is not prime.
- Forgetting that 1 is not a prime. The factorisation should only contain primes.
- Confusing factors with multiples. Factors divide the number exactly; multiples are built by multiplying the number.
- Missing a repeated prime in index form, for example writing \( 2 \times 5 \) instead of \( 2^2 \times 5 \) for 20.
- Assuming a number is prime without checking. Test divisibility by 2, 3, 5, 7, 11, 13 up to \( \sqrt{n} \).