Squaring a sum and a difference

A square of side 60 has area 3600; a square of side 5 has area 25. Can we use those to find the area of a square of side 65?

Draw the 65-square and cut it at 60 along both edges. It falls into four pieces: a 60 by 60 square, a 5 by 5 square, and two rectangles measuring 60 by 5.

So 65² = (60 + 5)² = 60² + 5² + 2 × (60 × 5) = 3600 + 25 + 600 = 4225.

Algebra gives the same four pieces: (60 + 5)(60 + 5) = 60 × 60 + 5 × 60 + 60 × 5 + 5 × 5 = 60² + 2 × (60 × 5) + 5².

Identity 1A: (a + b)² = a² + 2ab + b². Notice what it does not say: it isn't a² + b². The two rectangles in the middle are easy to forget, and easy to see in the picture.

Use it: (6x + 5)² = (6x)² + 2(6x × 5) + 5² = 36x² + 60x + 25. If you ever doubt the identity, just distribute instead (the book says so outright); the answer is the same.

Now the difference. Can 60² and 5² give us 55²? Draw a 55-square inside a 60-square. Start with 3600 and take away the two 60-by-5 strips along two edges. But those strips overlap in a 5 by 5 corner, so that corner has been taken away twice.

Add it back once: (60 − 5)² = 60² − (60 × 5) − (5 × 60) + 5² = 3600 − 300 − 300 + 25 = 3025.

Identity 1B: (a − b)² = a² + b² − 2ab. You can also get it from 1A by writing (a − b)² = (a + (−b))² = a² + (−b)² + 2 × a × (−b).

Watch the signs when you use it: 37² = (40 − 3)² = 1600 − 240 + 9, and −240 + 9 = −231 (not −249), so 37² = 1369.