June 2022 Paper 1 Q19
19 Write \(\quad 6x^2 - 24x + 17 \quad\) in the form \(\quad a(x + b)^2 + c \quad\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(6(x^2 - 4x \ldots\ldots\) or \(6(x - 2)^2 \ldots\ldots\) | M1 | oe eg \(6[(x^2 - 4x) \ldots\ldots]\) |
| \(6[(x - 2)^2 - 2^2] \ldots\ldots\) or \(6[(x - 2)^2 - 4] \ldots\ldots\) or \(6\left[(x - 2)^2 - 4 + \dfrac{17}{6}\right]\) or \(6\left[(x - 2)^2 - \dfrac{7}{6}\right]\) or \(6(x - 2)^2 - 6 \times \dfrac{7}{6}\) or \(6(x - 2)^2 - 24 + 17\) | M1dep | oe the bracket is after the \(2^2\) and the 4 here. If they put something else inside the bracket it is incorrect unless it is equivalent to one of the fully complete versions listed |
| \(6(x - 2)^2 - 7\) | A1 | |
| Alternative method 2 | ||
| \(ax^2 + 2abx + ab^2\;(+c)\) | M1 | expansion of brackets |
| \(a = 6\) and \(2ab = -24\) and \(ab^2 + c = 17\) | M1dep | |
| \(b = -2\) and \(c = -7\) | A1 | |