Let us use this identity to find the square of a number, say 119:

Step-by-step solution

Idea: “This identity” is (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. Split 119 by place value into hundreds, tens and ones, so every square and product is easy to do in your head.

1. 119 = 100 + 10 + 9, so a = 100, b = 10, c = 9.½ mark
2. 119² = 100² + 10² + 9² + 2(100)(10) + 2(100)(9) + 2(10)(9)½ mark
3. = 10000 + 100 + 81 + 2000 + 1800 + 180½ mark
4. Add: 10000 + 2000 + 1800 = 13800; 100 + 81 + 180 = 361; 13800 + 361 = 14161½ mark

Check: Using two terms: (120 − 1)² = 14400 − 240 + 1 = 14161 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Leaving out one of the three products, usually 2(10)(9) = 180. There must be three cross terms.
• Writing 2(100)(9) as 900 instead of 1800 (forgetting the 2).
• Adding the six numbers carelessly. Group the big ones and the small ones first.

How this can come in the exam

Try one yourself

Find 125² using 125 = 100 + 20 + 5.

More questions like this

• Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
• Factor using suitable identities:
• Expand the following using the identity
  (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
• Is this an identity?
  (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c².
• Look at the following figure. Justify the identity a² = (a + b) (a − b) + b² for yourself.

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