Foundation November 2022 Paper 1 Q26
26.
(a) Expand and simplify \((3x - 4)(5x + 7)\). [3]
(b)
Factorise each of the following.
(i) \(2x^2y + 12xy^2\) [3]
(ii) \(x^2 - 64\) [1]
| Scheme | Marks | Comments |
|---|---|---|
| \(15x^2 + 21x - 20x - 28\) | B2 | B1 for any three terms correct. \(mx^2 + x + n\) implies middle two terms correct if not from wrong working |
| \(15x^2 + x - 28\) | B1 | Mark final answer. Implies previous B2. FT their expression, provided it is a quadratic with 4 terms to consider and there are like terms to collect. |
| Scheme | Marks | Comments |
|---|---|---|
| (i) \(2xy(x + 6y)\) | B3 | Mark final answer. B2 for any one of the following:
|
| (ii) \((x - 8)(x + 8)\) | B1 | |
| (7) |
Corrected by Westie's Workshop
- The board's mark scheme printed “\(xy(2x + my)\) where \(m \neq 0\) or \(m \neq 6\)”. Changed to “\(xy(2x + my)\) where \(m \neq 0\) or \(m \neq 12\)” because \(xy(2x + 12y)\) is the correct answer (worth B2), so the value to leave out here is \(m = 12\), not 6.
- The board's mark scheme printed “\(xy(nx + 12y)\) where \(n \neq 1\) or \(n \neq 0\)”. Changed to “\(xy(nx + 12y)\) where \(n \neq 2\) or \(n \neq 0\)” because \(xy(2x + 12y)\) is the correct answer (worth B2), so the value to leave out here is \(n = 2\), not 1.
Original: see the board's mark scheme PDF (link above).