Higher June 2018 Paper 6 Q19
19 Show that \(\dfrac{2x^2 + 13x + 20}{2x^2 + x - 10}\) simplifies to \(\dfrac{x + a}{x - b}\) where \(a\) and \(b\) are integers. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{(2x + 5)(x + 4)}{(2x + 5)(x - 2)} = \dfrac{x + 4}{x - 2}\) | 4 | M3 for \((2x + 5)(x + 4)\) and \((2x + 5)(x - 2)\) seen OR M2 for \((2x + 5)(x + 4)\) or \((2x + 5)(x - 2)\) seen OR M1 for any two linear factors giving two correct terms in numerator or denominator | Warning: \(\dfrac{2(x + 5)(x + 4)}{2(x + 5)(x - 2)} = \dfrac{x + 4}{x - 2}\) scores SC1 eg. \((2x + 10)(x + 2)\) which gives \(2x^2\) and 20 |
| Alternative: M1 for \((2x^2 + 13x + 20)(x - b)\) and \((2x^2 + x - 10)(x + a)\) seen M1 two correct from \(-10a = -20b\) oe \(a - 10 = 20 - 13b\) oe \(2a + 1 = 13 - 2b\) oe M1dep (on M1M1) valid attempt to solve their simultaneous equations (condone one error) If 0 scored, allow SC2 for \(\dfrac{x + 4}{x - 2}\) as final answer from incomplete working, or SC1 for \(\dfrac{x + 4}{x - 2}\) seen. | |||