How can a factor tree be used to simplify surds?

√72 =
√2√36 =
√2√2√18 =
√2√2√2√9 =
√2√2√2(3) =
2√2(3) =
6√2

Example: √72

Step 1: Find the prime factorization of the number inside the radical.
√72
= √(2 * 2 * 2 * 3 * 3)

Step 2: Find pairs of the same number and move each repeated number from inside the radical to outside the radical. In this case, the pair of 2s and 3s moved outside the radical. The unpaired 2 stays inside the radical.
2 * 3 * √2

Step 3: Simplify the expressions both inside and outside the radical by multiplying.
6√2

Example: √1800

Step 1: Find the prime factorization of the number inside the radical.
√1800
= √(2 * 2 * 2 * 3 * 3 * 5 * 5)

Step 2: Find pairs of the same number and move each repeated number from inside the radical to outside the radical. In this case, the pair of 2s, 3s and 5s moved outside the radical. The unpaired 2 stays inside the radical.
2 * 3 * 5 * √2

Step 3: Simplify the expressions both inside and outside the radical by multiplying.
30√2

P.S. It's not necessary to factor the number all the way down to the prime factors. If you can find squared numbers, that will work too.
√1800
= √9 * √100 * √2
= 3 * 10 * √2
= 30√2

√1800 = √(2³×3²×5²)
= 2×3×5√2

If you use factor tree, you will have

72 = 2^3 x 3^2

So √72 = √(2^3 x 3^2)

= √2^3 x √3^2 ..............using √ab = √a x √b

= √(2^2 x 2) x √3^2

= √2^2 x √2 x √3^2

= 2 x √2 x 3

= 6√2

1800 = 2^3 x 3^2 x 5^2 ................using factor tree

√1800 = √(2^3 x 3^2 x 5^2)

= √2^3 x √3^2 x √5^2

= √(2^2 x 2) x √3^2 √5^2

= √2^2 x √2 x √3^2 x √5^2

= 2 x √2 x 3 x 5

= 30√2

√ 1800 = 10 √ 18 = 30 √2

√ 72 = 6 √2

Any factor that appears twice is a root.
For example, √8 =√2√2√2 = 2√2

√1800 = √18√100 = 10√18 = 10√9√2 = 10*3√2 = 30√2
√72 = √2√36 = 6√2