June 2025 Paper 2 Q8
8 \(k\) is a positive integer.
Prove that \(\quad 4k(k + 8) - (4k + 9)(k - 1) \quad\) is a multiple of 9 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4k^2 + 32k\) or \(4k^2 - 4k + 9k - 9\) or \(4k^2 + 5k - 9\) | M1 | may be seen in a full expansion eg \(\ldots - 4k^2 + 4k - 9k + 9\) may be seen in a grid |
| \(4k^2 + 32k - 4k^2 + 4k - 9k + 9\) or \(4k^2 + 32k - (4k^2 - 4k + 9k - 9)\) or \(4k^2 + 32k - 4k^2 - 5k + 9\) or \(4k^2 + 32k - (4k^2 + 5k - 9)\) | M1dep | correct 6-term expression and no errors or correct 5-term expression and no errors |
| M2 awarded and \(27k + 9\) and valid explanation involving 9 and \((3k + 1)\) | A1 | eg M2 awarded and \(27k + 9 = 9(3k + 1)\) or M2 awarded and \(\dfrac{27k + 9}{9} = 3k + 1\) |
Additional guidance
| \(4k^2 - 4k + 9k - 9\) followed by an error in simplifying to 3 terms eg \(4k^2 - 4k + 9k - 9 = 4k^2 - 5k - 9 \quad (-5k\) is an error) Award 1st M1 but no further marks can be awarded | M1M0A0 |
| Do not award 2nd M1 if errors seen eg \(4k(k + 8) = 4k^2 + 32k\) \((4k + 9)(k - 1) = 4k^2 + 4k - 9k - 9 = 4k^2 + 5k - 9 \quad (+4k\) and \(-9k\) are errors) \(4k^2 + 32k - 4k^2 - 5k + 9\) (‘correct’ 5-term expression with errors) | M1 M0A0 |
| Missing brackets cannot be recovered | |
| For M marks accept eg \(- -9\) for \(+9\) | |
| Valid explanations involving 9 and \((3k + 1)\) include eg1 \(27k + 9\) when divided by 9 is \((3k + 1)\) eg2 \(27k + 9\) has factors \((3k + 1)\) and 9 | |
| Incomplete explanations include eg1 \(27k + 9\) which is divisible by 9 eg2 \(27k + 9\) and \(27k \div 9 = 3k\) and \(9 \div 9 = 1\) | |
| Condone eg \(x\) used for \(k\) |