Higher June 2025 Paper 2 Q23
23 \(\dfrac{4x^2 - 4x - 120}{5x^2 - 180} \div \dfrac{x^2 + 5x}{10x^2 + 60x} = p\) where \(p\) is an integer.
Find the value of \(p\)
Show clear algebraic working.
(4)
| Scheme | Marks |
|---|---|
\(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \div \dfrac{x(x + 5)}{10x(x + 6)}\;(= p)\) or \(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}\;(= p)\) or 2 from \(4(x - 6)(x + 5)\) or \(5(x - 6)(x + 6)\) or \(x(x + 5)\) or \(10x(x + 6)\) | M1 |
\(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \div \dfrac{x(x + 5)}{10x(x + 6)}\;(= p)\) or \(\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}\;(= p)\) or \(4(x - 6)(x + 5)\) and \(5(x - 6)(x + 6)\) and \(x(x + 5)\) and \(10x(x + 6)\) | M1 |
| \(\dfrac{4x^2 - 4x - 120}{5x^2 - 180} \times \dfrac{10x^2 + 60x}{x^2 + 5x}\;(= p)\) | M1 |
| Working required Answer: 8 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for factorising 2 or 3 of the quadratics fully – could be implied by 2 factors cancelled correctly
NB factors must be in the form \((ax + b)\)
NB Substitution of values of \(x\) into the given equation is not an acceptable algebraic method
M1: for factorising all of the quadratics fully – could be implied by 2 factors cancelled correctly
NB factors must be in the form \((ax + b)\)
M1: for inverting the 2nd fraction (this mark can be awarded at any time and may be awarded with incorrect factorisation if meaning is clear)
A1: oe dep on M3
ALT
| Scheme | Marks |
|---|---|
| \(\dfrac{40x^4 + 200x^3 - 1440x^2 - 7200x}{5x^4 + 25x^3 - 180x^2 - 900x}\) | M3 |
| 8 | A1 |
Notes
M3: for a correct expression (no errors)
A1: oe dep on M3