Higher June 2025 Paper 2R Q13
13
(a) Factorise \(4x^2 - 25y^2\) (2)
(b) Show that \(4x(x + 3)(2x - 5)\) can be written in the form \(ax^3 + bx^2 + cx\) where \(a\), \(b\) and \(c\) are integers to be found. (3)
| Scheme | Marks |
|---|---|
| \((2x \pm 5y)(2x \pm 5y)\) or \((2x)^2 - (5y)^2\) | M1 |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: \((2x + 5y)(2x - 5y)\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(4x(x + 3) = 4x^2 + 12x\) or \(4x(2x - 5) = 8x^2 - 20x\) or \((x + 3)(2x - 5) = 2x^2 - 5x + 6x - 15\;(= 2x^2 + x - 15)\) | M1 |
| \((4x^2 + 12x)(2x - 5) = 8x^3 - 20x^2 + 24x^2 - 60x\) or \((8x^2 - 20x)(x + 3) = 8x^3 + 24x^2 - 20x^2 - 60x\) or \(4x(2x^2 - 5x + 6x - 15) = 8x^3 - 20x^2 + 24x^2 - 60x\) or \(4x(2x^2 + x - 15) = 8x^3 + 4x^2 - 60x\) | M1 |
| Working required Answer: \(8x^3 + 4x^2 - 60x\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: An expansion with only one error
Do not award this mark for
\(4x^2 + 12x + 8x^2 - 20x\)
M1: ft dep on M1
allow one further error
A1: cao dep on M1
Terms may be in any order but must be simplified
ISW correct factorisation
\(8x^3 + 4x^2 - 60x\) must be seen previously to award 3 marks
eg \(4(2x^3 + x^2 - 15x)\)
\(x(8x^2 + 4x - 60)\)
Do not ISW incorrect simplification or further incorrect work following \(8x^3 + 4x^2 - 60x\)
eg \(8x^3 + 4x^2 - 60x = 2x^3 + x^2 - 15x\) gets M2A0
M2 for 3 terms (out of a maximum of 4 terms) from:
\(8x^3 - 20x^2 + 24x^2 - 60x\)
If not M2, then M1 for 2 correct out of a maximum of 4