Higher June 2019 Paper 1 Q16
16 Simplify fully \(\qquad \dfrac{4x - 8x^2}{12x - 6}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Correct factorisation of numerator \(2(2x - 4x^2)\) or \(4(x - 2x^2)\) or \(x(4 - 8x)\) or \(2x(2 - 4x)\) or \(4x(1 - 2x)\) or correct factorisation of denominator \(2(6x - 3)\) or \(3(4x - 2)\) or \(6(2x - 1)\) or correct cancelling by 2 throughout \(\dfrac{2x - 4x^2}{6x - 3}\) | M1 | oe with negative coefficients |
| Correct fraction with numerator \(4x(1 - 2x)\) or \(-4x(2x - 1)\) and denominator \(6(2x - 1)\) or \(-6(1 - 2x)\) or \(-\dfrac{4x}{6}\) or \(\dfrac{-4x}{6}\) or \(\dfrac{4x}{-6}\) or \(\dfrac{2x(2 - 4x)}{-3(2 - 4x)}\) or \(\dfrac{2x(2 - 4x)}{3(4x - 2)}\) | M1dep | oe with cancelling of 2 throughout eg \(\dfrac{2x(1 - 2x)}{3(2x - 1)}\) or \(\dfrac{2x(1 - 2x)}{-3(1 - 2x)}\) |
| \(-\dfrac{2x}{3}\) or \(-\dfrac{2}{3}x\) | A1 | allow \(\dfrac{-2x}{3}\) or \(\dfrac{2x}{-3}\) |
Additional guidance
Allow multiplication signs up to M1M1
Allow \(-0.\dot{6}\) for \(-\dfrac{2}{3}\)
Do not allow \(-0.66\ldots\) for \(-\dfrac{2}{3}\)
For the first M1 only, allow any correct factorisation seen within multiple attempts