June 2025 Paper 1 Q14
14 Simplify fully \(\qquad \dfrac{2x^2 - 4x - 6}{x^2 - 3x} \div \dfrac{4x + 4}{7x^2}\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2(x - 3)(x + 1)\) or \((2x - 6)(x + 1)\) or \((x - 3)(2x + 2)\) | M1 | |
| \(x(x - 3)\) or \(4(x + 1)\) or \(2(2x + 2)\) | M1 | oe for either of the other expressions factorised |
| \(\dfrac{2(x - 3)(x + 1)}{x(x - 3)} \times \dfrac{7x^2}{4(x + 1)}\) | M1dep | oe depends on both previous M marks and must all be correctly factorised into something that can be fully cancelled could imply previous M marks |
| \(\dfrac{7}{2}x\) or \(3.5x\) or \(3\dfrac{1}{2}x\) | A1 | |
| Alternative method 2: cancelled factor of 2 first | ||
| \(\dfrac{x^2 - 2x - 3}{x^2 - 3x} \div \dfrac{2x + 2}{7x^2}\) | M1 | oe where 2 has been cancelled |
| \((x - 3)(x + 1)\) or \(x(x - 3)\) or \(2(x + 1)\) | M1 | oe for any expression factorised correctly |
| \(\dfrac{(x - 3)(x + 1)}{x(x - 3)} \times \dfrac{7x^2}{2(x + 1)}\) | M1dep | oe depends on both previous M marks and must all be correctly factorised into something that can be fully cancelled could imply previous M marks |
| \(\dfrac{7}{2}x\) or \(3.5x\) or \(3\dfrac{1}{2}x\) | A1 | |
| Alternative method 3: multiplying first | ||
| \(\dfrac{7x^2\left(2x^2 - 4x - 6\right)}{\left(x^2 - 3x\right)\left(4x + 4\right)}\) | M1 | oe eg \(\dfrac{14x^4 - 28x^3 - 42x^2}{4x^3 - 8x^2 - 12x}\) where fractions have been multiplied and combined into a single fraction |
| Numerator or Denominator further factorised correctly as a step towards the final answer | M1 | eg \(14x^2\left(x^2 - 2x - 3\right)\) or \(14x^2(x - 3)(x + 1)\) or \(2x\left(2x^2 - 4x - 6\right)\) or \(4x\left(x^2 - 2x - 3\right)\) or \(4x(x - 3)(x + 1)\) etc could be implied by cancellation of a common factor neither \(7x^2\left(2x^2 - 4x - 6\right)\) nor \(\left(x^2 - 3x\right)\left(4x + 4\right)\) would score this M mark |
| \(\dfrac{7x^2\left(2x^2 - 4x - 6\right)}{2x\left(2x^2 - 4x - 6\right)}\) or \(\dfrac{14x^2\left(x^2 - 2x - 3\right)}{4x\left(x^2 - 2x - 3\right)}\) | M1dep | oe depends on both previous M marks and must all be correctly factorised into something that can be fully cancelled |
| \(\dfrac{7}{2}x\) or \(3.5x\) or \(3\dfrac{1}{2}x\) | A1 | |
Additional guidance
| Condone factorising outside of the fractions for first two M marks for Alt 1 and Alt 2 and on the second mark for Alt 3 | |
| Alt 2 it is possible to cancel with \(x\) or \(2x\). Mark as an equivalent to Alt 2 | |
| Follow whichever Alternative method gives the best mark |