Foundation June 2019 Paper 3 Q19
19
(a) Simplify fully \(\quad 3a^2 + 7a + 3 - a^2 + 8a - 4\) [3 marks]
(b) Factorise fully \(\quad 24y^2 - 20y\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2a^2 + 15a - 1\) | B3 | B2 \(2a^2 + 15a\) or \(2a^2 - 1\) or \(15a - 1\) B1 \(2a^2\) or \(15a\) or \(-1\) |
Additional guidance
| \(2a + 15a - 1 = 17a - 1\) | B2 |
| \(2a^2 + 15a + -1\) | B2 |
| Do not ignore further incorrect algebraic simplification for B3 \(2a^2 + 15a - 1 = 17a - 1\) | B2 |
| Do not ignore further incorrect algebraic simplification for B2 \(2a + 15a - 1 = 17a - 1 = 16a\) \(2a^2 + 15a - 1 = 17a - 1 = 16a\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(4y(6y - 5)\) or \(-4y(5 - 6y)\) | B2 | B1 \(2y(12y - 10)\) or \(-2y(10 - 12y)\) or \(y(24y - 20)\) or \(-y(20 - 24y)\) or \(4(6y^2 - 5y)\) or \(-4(5y - 6y^2)\) or \(2(12y^2 - 10y)\) or \(-2(10y - 12y^2)\) |
Additional guidance
| Ignore any ‘solutions’ seen eg \(4y(6y - 5)\) in working with 0 and \(\dfrac{5}{6}\) on answer line | B2 |
| Condone \(4y \times (6y - 5)\) | B2 |
| Condone \(y \times (24y - 20)\) | B1 |
| \((4y + 0)(6y - 5)\) | B1 |
| Do not ignore further incorrect algebraic simplification for B2 |