Higher November 2017 Paper 2 Q16
16
(a) Factorise fully \(\quad 9y^3 - 6y\) [2 marks]
(b) Factorise \(\quad 3x^2 - 22x + 7\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3y(3y^2 - 2)\) or \(-3y(2 - 3y^2)\) | B2 | B1 \(3(3y^3 - 2y)\) or \(y(9y^2 - 6)\) or \(-3(2y - 3y^3)\) or \(-y(6 - 9y^2)\) |
Additional guidance
| \(3y(3y^2 - 2)\) or \(-3y(2 - 3y^2)\) followed by incorrect further work eg \(3y(3y^2 - 2) = 3y^2(3y - 2)\) | B1 |
| \(3y(3y^2 - 2) = 3y(\sqrt{3}y + 2)(\sqrt{3}y - 2)\) | B2 |
| \(3y(3y^2 - 2) = 9y^3 - 6y\) (checking) | B2 |
| \(3y \times (3y^2 - 2)\) | B2 |
| \(3 \times (3y^3 - 2y)\) | B1 |
| \(y3(3y^2 - 2)\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \((3x - 1)(x - 7)\) or \((1 - 3x)(7 - x)\) | B2 | B1 \((3x + a)(x + b)\) where \(ab = 7\) or \(a + 3b = -22\) or \((a - 3x)(b - x)\) where \(ab = 7\) or \(a + 3b = 22\) |
Additional guidance
| \((3x + 1)(x + 7)\) | B1 |
| \((3x - 1)(x - 7)\) | B1 |
| \((3x - 4)(x - 6)\) | B1 |
| \((7 - 3x)(1 - x)\) | B1 |
| \((10 - 3x)(4 - x)\) | B1 |
| \((3x - 1) \times (x - 7)\) | B2 |
| Ignore any ‘solutions’ seen eg \((3x - 1)(x - 7)\) in working with \(\dfrac{1}{3}\) and 7 on answer line | B2 |
Notes
The second guidance row is printed this way in the published mark scheme. The correct answer \((3x - 1)(x - 7)\) scores B2, as shown in the main table.