Higher November 2023 Paper 6 Q21
21 Simplify fully.
\[\frac{x^3 + 8x^2 + 15x}{x^3 - 9x}\][5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x + 5}{x - 3}\) | 5 | M4 for \(\frac{x(x + 3)(x + 5)}{x(x + 3)(x - 3)}\) or \(\frac{(x + 3)(x + 5)}{(x + 3)(x - 3)}\) OR M2 for \([x](x + 3)(x + 5)\) or M1 for \([x](x(x + 5) + 3(x + 5))\) or \([x](x(x + 3) + 5(x + 3))\) or for \([x](x + a)(x + b)\) where \(a + b = 8\) or \(ab = 15\) and M1 for \([x](x + 3)(x - 3)\) or \(\frac{x(x^2 + 8x + 15)}{x(x^2 - 9)}\) or \(\frac{x^2 + 8x + 15}{x^2 - 9}\) or for \(x^2 + 8x + 15\) and \(x^2 - 9\) or any other partially factorised form of the numerator or denominator | For M2 and M1 marks, if written as a quotient, condone [\(x\)] not being consistently present in both or cancelled out from both Also award M2 and M1 marks for factorising without first factorising [\(x\)]. eg. \((x^2 + 3x)(x + 5)\) earns M2 |