June 2019 Paper 2 Q9
9 Factorise fully \(\quad (p + 6)^{11} - (p + 6)^{10}\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((p + 6)^{10}(p + 5)\) or \((p + 5)(p + 6)^{10}\) | B2 | B1 \((p + 6)^{10}(p + 6 - 1)\) or states \(x = p + 6\) and \(x^{10}(x - 1)\) (any letter for \(x\) other than \(p\)) or correct partial factorisation eg \((p + 6)[(p + 6)^{10} - (p + 6)^9]\) or \((p + 6)^2[(p + 6)^9 - (p + 6)^8]\) |
Additional guidance
| Any shape of bracket may be used | |
| \((p + 6)^{10}((p + 6) - 1)\) | B1 |
| Missing brackets must be recovered eg \(\ p + 5\,(p + 6)^{10}\) not recovered and B1 response not seen | B0 |
| Condone \(\ (p + 6)^{10}(p + 5\) | B2 |
| Condone \(\ (p + 6)^{10}(p + 6 - 1\) | B1 |
| \((p + 6)^{10}(p + 5)\) followed by expansion attempt | B1 |
| B1 response followed by expansion attempt | B1 |
| \((p + 6)^{10} \times (p + 5)\) | B1 |
| Condone multiplication signs for B1 eg \(\ (p + 6)^{10} \times (p + 6 - 1)\) | B1 |
| \((p + 6)^{11}\left[1 - \dfrac{1}{p + 6}\right]\) | B1 |