June 2024 Paper 1 Q17
17 Show that \(\quad \dfrac{21x}{3x^2 - 2x - 8} - \dfrac{7}{x - 2} \quad\) simplifies to \(\quad \dfrac{k}{3x^2 - 2x - 8} \quad\) where \(k\) is an integer. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((3x + 4)(x - 2)\) | M1 | oe for correct factorisation |
| \(\dfrac{21x - 7(3x + 4)}{(3x + 4)(x - 2)}\) | M1dep | oe fractions can be written separately denominator must be \((3x + 4)(x - 2)\) or \(3x^2 - 2x - 8\) |
| \(\dfrac{-28}{3x^2 - 2x - 8}\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{21x(x - 2) - 7(3x^2 - 2x - 8)}{(3x^2 - 2x - 8)(x - 2)}\) | M1 | oe fractions can be written separately |
| \(\dfrac{21x^2 - 42x - 21x^2 + 14x + 56}{(3x^2 - 2x - 8)(x - 2)}\) | M1dep | oe must be one fraction and brackets expanded in the numerator (could have been simplified and factorised) eg \(\dfrac{-28(x - 2)}{(3x^2 - 2x - 8)(x - 2)}\) expanded denominator is \(3x^3 - 8x^2 - 4x + 16\) |
| \(\dfrac{-28}{3x^2 - 2x - 8}\) | A1 | |
Additional guidance
| Students may use a combination of methods which are covered by oe | |
| \(k = -28\) Could come from using identities or substitutions | SC1 |
| The minus can be written before the fraction rather than with the 28 |