Suppose you have to simplify \sqrt{72} mid-exam, with four sub-parts still to go. Two roads.
Road one — prime factorisation. 72 = 2^3 \cdot 3^2. Group into pairs: one pair of 2s, one pair of 3s, one leftover 2. Pairs escape as 2 \cdot 3 = 6. Loner 2 stays. Answer: \sqrt{72} = 6\sqrt{2}.
Road two — spot the biggest perfect square. You know 36 is a perfect square and 72 = 36 \cdot 2. So \sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}.
Same answer. Road one took four arithmetic operations; road two took one recognition and one multiplication. Under time pressure, road two wins.
This article is the habit: memorise the perfect squares, train your eye to spot them as factors, and save prime factorisation for the cases where recognition fails.
The perfect squares you should memorise
If you do not already know these cold, learn them now. Every one of them will earn you seconds on an exam paper.
These are 1^2 through 25^2. For class 10 and class 12 boards, knowing up to 400 (20^2) is enough. For JEE Main, knowing up to 625 (25^2) covers almost every radical you will meet. For olympiads, keep going: 676, 729, 784, 841, 900, 961, 1024, \ldots
You do not memorise these as a separate table. You memorise them by using them — every time you square something or spot a perfect-square factor, that number lodges a little deeper.
The recognition move
Here is the drill. For \sqrt{n}, scan your memorised list and ask: what is the largest perfect square that divides n?
- \sqrt{50}: is 25 a factor of 50? Yes, 50 = 25 \cdot 2. So \sqrt{50} = 5\sqrt{2}.
- \sqrt{98}: is 49 a factor of 98? Yes, 98 = 49 \cdot 2. So \sqrt{98} = 7\sqrt{2}.
- \sqrt{162}: is 81 a factor of 162? Yes, 162 = 81 \cdot 2. So \sqrt{162} = 9\sqrt{2}.
- \sqrt{288}: is 144 a factor of 288? Yes, 288 = 144 \cdot 2. So \sqrt{288} = 12\sqrt{2}.
- \sqrt{507}: is 169 a factor of 507? 507 / 169 = 3. Yes. So \sqrt{507} = 13\sqrt{3}.
Notice what you did not do. You did not factor 288 into 2^5 \cdot 3^2, pair up the primes, and extract. You looked at 288, saw "that is 144 \cdot 2," and wrote the answer.
Why this is faster than prime factorisation for exam problems
Prime factorisation is a reliable algorithm — it never fails. But it is a process: divide by 2, divide again, divide by 3, list the prime powers, group into pairs. For 288, that is five or six arithmetic steps.
Perfect-square recognition is one check: is this divisible by the square I suspect? If yes, divide once and you are done.
The brain running a search beats the brain running an algorithm. You have three hours, ninety questions, and every radical simplification is a micro-decision. Recognition wins.
When prime factorisation wins
Not every number hands its perfect-square factor to you for free. When n has many distinct prime factors, when n is given in prime-power form already (like \sqrt{2^{10} \cdot 3^7 \cdot 5^3}), or when n is unusually large — prime factorisation is the right tool. For \sqrt{2^{10} \cdot 3^7 \cdot 5^3} you pair exponents directly: 2^5 \cdot 3^3 \cdot 5 \sqrt{3 \cdot 5} = 4320\sqrt{15}. That is the method from the prime-factor jailbreak. Use the right tool for the shape of the number.
The divisibility tricks for finding perfect-square factors
You do not have to guess whether 4, 9, 25 divide n.
- Divisible by 4 iff the last two digits form a number divisible by 4. 288: last two digits 88 = 4 \cdot 22. Yes.
- Divisible by 9 iff the digit sum is divisible by 9. 162: digit sum 9. Yes.
- Divisible by 25 iff the number ends in 00, 25, 50, or 75.
- Divisible by 16 iff divisible by 4 twice (easier than checking the last four digits).
- Divisible by 36 iff divisible by both 4 and 9.
- Divisible by 100 iff ends in 00.
For \sqrt{2700}: digit sum 9, ends in 00, so divisible by 900. And 2700 / 900 = 3. Answer: \sqrt{2700} = 30\sqrt{3}. Three seconds of mental arithmetic.
Generalisation to higher roots
The same habit carries over to cube roots — and fourth roots, fifth roots, any root. For \sqrt[3]{n}, spot the biggest perfect cube factor.
- \sqrt[3]{54}: is 27 a factor? 54 = 27 \cdot 2. So \sqrt[3]{54} = 3\sqrt[3]{2}.
- \sqrt[3]{128}: is 64 a factor? 128 = 64 \cdot 2. So \sqrt[3]{128} = 4\sqrt[3]{2}.
- \sqrt[3]{250}: is 125 a factor? 250 = 125 \cdot 2. So \sqrt[3]{250} = 5\sqrt[3]{2}.
The perfect cubes to know: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Ten numbers. Memorise them; they cover every cube root you will meet below \sqrt[3]{2000}.
The hybrid — peel layer by layer
Sometimes the biggest perfect square is not the first one you spot. Peel.
\sqrt{200}: you first see that 4 divides 200, so \sqrt{200} = 2\sqrt{50}. Now \sqrt{50}: 25 divides 50, so \sqrt{50} = 5\sqrt{2}. Combine: \sqrt{200} = 10\sqrt{2}.
Or you spot 100 directly: 200 = 100 \cdot 2, so \sqrt{200} = 10\sqrt{2}. Same answer, one step instead of two. If you do not see the biggest square, pull out any square and keep peeling until nothing more comes out.
Sanity check — always get the same answer
Whichever method you use, verify by squaring the result.
- 6\sqrt{2} squared: 6^2 \cdot 2 = 36 \cdot 2 = 72. Matches \sqrt{72}. ✓
- 5\sqrt{2} squared: 25 \cdot 2 = 50. Matches \sqrt{50}. ✓
- 10\sqrt{2} squared: 100 \cdot 2 = 200. Matches \sqrt{200}. ✓
If the squared answer does not reproduce the original n, you made an arithmetic slip somewhere. This check takes two seconds and catches almost every common mistake.
Worked drill — do these in your head
Cover the right column. Work each one before you read on.
- \sqrt{75}: 25 \cdot 3, so 5\sqrt{3}.
- \sqrt{48}: 16 \cdot 3, so 4\sqrt{3}.
- \sqrt{300}: 100 \cdot 3, so 10\sqrt{3}.
- \sqrt{1200}: 400 \cdot 3, so 20\sqrt{3}.
- \sqrt{63}: 9 \cdot 7, so 3\sqrt{7}. (Do not reach for 49 — it does not divide 63.)
- \sqrt{242}: 121 \cdot 2, so 11\sqrt{2}.
- \sqrt{175}: 25 \cdot 7, so 5\sqrt{7}.
- \sqrt{108}: 36 \cdot 3, so 6\sqrt{3}.
- \sqrt{363}: 121 \cdot 3, so 11\sqrt{3}.
- \sqrt{500}: 100 \cdot 5, so 10\sqrt{5}.
If you got nine or ten out of ten without writing anything down, your recognition is working. If you got fewer, the perfect-squares list needs another pass.
When you cannot spot anything
You will meet numbers where no perfect square jumps out. \sqrt{231}, for example: 231 is odd, digit sum 6 (not divisible by 9), does not end in 25 or 75. No 4, 9, or 25 factor. You try 49: 231 / 49 is not an integer. Then 121: no. At that point, default to the reliable method — prime-factor jailbreak. Factorise: 231 = 3 \cdot 7 \cdot 11. All three primes are singletons, nothing pairs, \sqrt{231} is already in simplest form.
The lesson: recognition is your first line, not your only line. If nothing comes up in a few seconds of scanning, switch tools and factorise. Do not burn a minute hunting for a square that is not there.
Closing
Memorise the perfect squares to 625, and the perfect cubes to 1000. When a radical shows up, scan: is the number divisible by 4? 9? 25? 36? Pull out the biggest square you recognise, write the simplified form, square-check, move on. Save the full factor tree for the cases where nothing catches your eye. That is not laziness — it is exam technique. The person next to you is still drawing factor trees for \sqrt{72}; you have already written 6\sqrt{2} and started the next question.