June 2017 Paper 2 Q12
12
(a) Factorise fully \(\qquad 75 - 3x^2\) [2 marks]
(b) Simplify fully \(\qquad (3n + 1)^2 - (3n - 1)^2\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3(25 - x^2)\) or \(-3(x^2 - 25)\) or \((15 + 3x)(5 - x)\) or \((x + 5)(15 - 3x)\) | M1 | oe partial factorisation eg \(-(3x + 15)(x - 5)\) Brackets in either order Do not allow \(-(3x^2 - 75)\) |
| \(3(5 + x)(5 - x)\) or \(3(-x - 5)(x - 5)\) or \(-3(x + 5)(x - 5)\) or \(-3(5 - x)(-x - 5)\) | A1 |
Additional guidance
| \((-x + 5)\) is equivalent to \((5 - x)\) etc | |
| Do not allow A1 for incorrect notation in final answer eg \((5 + x)3(5 - x)\) | M1A0 |
| Do not allow A1 for use of multiplication signs in final answer eg \(3 \times (5 + x) \times (5 - x)\) | M1A0 |
| Correct answer followed by incorrect further work | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(9n^2 + 3n + 3n + 1\) or \(9n^2 + 6n + 1\) or \(9n^2 - 3n - 3n + 1\) or \(9n^2 - 6n + 1\) | M1 | oe Terms may be seen in a grid |
| \(12n\) with no incorrect working | A1 | Brackets can be recovered |
| Alternative method 2 | ||
| \((3n + 1 + 3n - 1)(3n + 1 - (3n - 1))\) or \((3n + 1 + 3n - 1)(3n + 1 - 3n + 1)\) | M1 | oe Brackets around \(3n - 1\) can be recovered |
| \(12n\) | A1 | |
Additional guidance
| Alt 1 \(12n\) may come from incorrect working eg1 \(3n^2 + 6n + 1 - (3n^2 - 6n + 1) = 12n\) eg2 \(9n^2 + 3n + 1 - (9n^2 - 3n + 1) = 12n\) | M0A0 M0A0 |
| Alt 1 Recovery of brackets eg1 \(9n^2 + 6n + 1 - 9n^2 - 6n + 1 = 12n\) eg2 \(9n^2 + 6n + 1 - 9n^2 - 6n + 1 = 2\) | M1A1 M1A0 |
| Alt 2 Recovery of brackets eg1 \((3n + 1 + 3n - 1)(3n + 1 - 3n - 1) = 12n\) eg2 \((3n + 1 + 3n - 1)(3n + 1 - 3n - 1) = 0\) | M1A1 M0A0 |
| Do not allow A1 for use of multiplication signs in final answer eg \(12 \times n\) with no incorrect working | M1A0 |