Prime factorisation calculator with a factor tree

1. Split the number into two factors, for example $360 = 2 \times 180$.
2. Circle any factor that is prime (2). Keep splitting the others: $180 = 2 \times 90$, $90 = 2 \times 45$, $45 = 3 \times 15$, $15 = 3 \times 5$.
3. Stop when every branch ends in a prime.
4. Multiply the circled primes together, grouping repeats as powers: $2^3 \times 3^2 \times 5$.

The calculator always splits off the smallest prime first, which gives a neat tree, but any first split gives the same primes in the end. That's the fundamental theorem of arithmetic: every whole number above 1 has exactly one prime factorisation.

Repeated primes are written as powers: $2 \times 2 \times 2 = 2^3$. Exams usually ask for the answer "as a product of prime factors in index form", so $360 = 2^3 \times 3^2 \times 5$, not $2 \times 2 \times 2 \times 3 \times 3 \times 5$, unless the question says otherwise.

For more, see our guide to factor trees and the prime factorisation topic guide.

• Circling a number that isn't prime, such as 9 or 15. Keep splitting until every end is prime.
• Treating 1 as prime. It isn't, and it never appears in a factor tree's answer.
• Forgetting a branch when multiplying the primes back together. Check by multiplying out: you should get the original number.

A number is a perfect square when every power in its prime factorisation is even: $576 = 2^6 \times 3^2$, so $576 = (2^3 \times 3)^2 = 24^2$. It's a perfect cube when every power is a multiple of 3: $216 = 2^3 \times 3^3 = 6^3$.

This gives a neat exam technique. To find the smallest number to multiply 72 by to make a square, write $72 = 2^3 \times 3^2$. The power of 2 is odd, so multiply by one more 2: $72 \times 2 = 144 = 12^2$.