Higher November 2024 Paper 1 Q15
15 Write \(3x(2x - 1)(5x + 4)\) in the form \(ax^3 + bx^2 + cx\) where \(a\), \(b\) and \(c\) are integers.
(3)
| Scheme | Marks |
|---|---|
| \(3x(2x - 1) = 6x^2 - 3x\) or \(3x(5x + 4) = 15x^2 + 12x\) or \((2x - 1)(5x + 4) = 10x^2 + 8x - 5x - 4\) \((10x^2 + 3x - 4)\) | M1 |
| \((6x^2 - 3x)(5x + 4) = 30x^3 + 24x^2 - 15x^2 - 12x\) \((15x^2 + 12x)(2x - 1) = 30x^3 - 15x^2 + 24x^2 - 12x\) \(3x(10x^2 + 8x - 5x - 4) = 30x^3 + 24x^2 - 15x^2 - 12x\) \(3x(10x^2 + 3x - 4) = 30x^3 + 9x^2 - 12x\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(30x^3 + 9x^2 - 12x\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: An expansion with only one error.
Do not award this mark for
\(6x^2 - 3x + 15x^2 + 12x\)
M1: ft dep on M1
allow one further error
A1: cao (terms may be in any order but must be simplified) dep on M1
accept \(a = 30\), \(b = 9\), \(c = -12\)
ISW correct factorisation
eg
\(3(10x^3 + 3x^2 - 4x)\)
Do not ISW incorrect simplification
M2 for 3 (out of a maximum of 4) of
\(30x^3 + 24x^2 - 15x^2 - 12x\)
(M1 for 2 correct out of a maximum of 4)