Higher June 2025 Paper 2 Q12
12 Expand and simplify \(3x(2x + 5)(7x - 4)\)
(3)
| Scheme | Marks |
|---|---|
| \(3x(2x + 5) = 6x^2 + 15x\) or \(3x(7x - 4) = 21x^2 - 12x\) or \((2x + 5)(7x - 4) = 14x^2 - 8x + 35x - 20\) \((14x^2 + 27x - 20)\) | M1 |
| \((6x^2 + 15x)(7x - 4) = 42x^3 - 24x^2 + 105x^2 - 60x\) \((21x^2 - 12x)(2x + 5) = 42x^3 + 105x^2 - 24x^2 - 60x\) \(3x(14x^2 - 8x + 35x - 20) = 42x^3 - 24x^2 + 105x^2 - 60x\) \(3x(14x^2 + 27x - 20) = 42x^3 + 81x^2 - 60x\) | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(42x^3 + 81x^2 - 60x\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: An expansion with only one error.
Do not award this mark for
\(6x^2 + 15x + 21x^2 - 12x\)
or
\((6x^2 + 15x)(21x^2 - 12x)\)
M1: ft dep on M1
allow one further error
A1: cao (terms may be in any order but must be simplified) dep on M1
ISW correct factorisation eg \(3(14x^3 + 27x^2 - 20x)\)
Do not ISW incorrect simplification eg \(14x^3 + 27x^2 - 20x\)
M2 for 3 (out of a maximum of 4) of \(42x^3 + 105x^2 - 24x^2 - 60x\)
(M1 for 2 correct out of a maximum of 4)