Foundation January 2023 Paper 1R Q23
23
(a) Factorise fully
\(18c^3d^2 - 21c^2\) (2)
\(18c^3d^2 - 21c^2\) (2)
(b)
(i) Factorise
\(y^2 - 3y - 18\) (2)
\(y^2 - 3y - 18\) (2)
(ii) Hence, solve
\(y^2 - 3y - 18 = 0\) (1)
\(y^2 - 3y - 18 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(3c^2(6cd^2 - 7)\) | B2 |
| (2) |
Notes
B2: fully correct or
B1 for a correct partial factorisation with at least two terms outside the bracket ie \(3c(6c^2d^2 - 7c)\) or \(c^2(18cd^2 - 21)\)
or the fully correct factor outside the bracket with two terms inside the bracket and at most one mistake \(3c^2(\ldots\ldots)\)
| Scheme | Marks |
|---|---|
| (i) eg \((y \pm 6)(y \pm 3)\) or \(y(y + 3) - 6(y + 3)\) or \(y(y - 6) + 3(y - 6)\) | M1 |
| [allow use of \(x\) rather than \(y\)] Answer: \((y - 6)(y + 3)\) | A1 |
| (ii) Answer: 6, –3 | B1 |
| (3) | |
| (5 marks) |
Notes
M1: or \((y + a)(y + b)\) where \(ab = -18\) or \(a + b = -3\)
or
factorisation which expands to give 2 out of 3 correct terms
B1: ft must come from their factors in (b)(i)