4MA0 Higher January 2018 Paper 3 Q21
21 Factorise completely \((10a - b)^2 - (2a - 5b)^2\)
(3)
| Scheme | Marks |
|---|---|
| \(((10a - b) + (2a - 5b))((10a - b) - (2a - 5b))\) Or \((10a - b + 2a - 5b)(10a - b - 2a + 5b)\) | M1 |
| \((12a - 6b)(8a + 4b)\) | M1 |
| \(24(2a - b)(2a + b)\) | A1 |
| (3) |
Notes
M1: first stage in using difference of 2 squares
allow \(-2a - 5b\) in place of \(-(2a - 5b)\)
M1: simplifying 2 correct brackets
A1: fully factorised expression from correct working
Alternative
| Scheme | Marks |
|---|---|
| \((100a^2 - 10ab - 10ab + b^2)\) or \(4a^2 - 10ab - 10ab + 25b^2\) or \(-4a^2 + 10ab + 10ab - 25b^2\) | M1 |
| \(96a^2 - 24b^2\) | M1 |
| \(24(2a - b)(2a + b)\) | A1 |
Notes
M1: expanding one bracket correctly
M1: simplifying 2 correct expansions
A1: fully factorised expression from correct working