Higher June 2019 Paper 2R Q22
22 Write \(5 + 12x - 2x^2\) in the form \(a + b(x + c)^2\) where \(a\), \(b\) and \(c\) are integers.
(4)
| Scheme | Marks |
|---|---|
| \(-2(x^2 - 6x) + 5\) or \(-2(x^2 - 6x - 2.5)\) | M1 |
| \(-2[(x - 3)^2 - 9 - 2.5]\) or \(-2[(x - 3)^2 - 9] + 5\) | M1 |
| \(-2[(x - 3)^2 - 11.5]\) or \(-2(x - 3)^2 + 18 + 5\) | M1 |
| \(23 - 2(x - 3)^2\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Factorising by extracting −2 in a correct expression
M1: Correct expression equivalent to \(5 + 12x - 2x^2\)
M1: Correct expression equivalent to \(5 + 12x - 2x^2\)
A1: Award full marks if \(a\), \(b\), and \(c\) are correctly stated and \(23 - 2(x - 3)^2\) is not stated anywhere.
SC B3 for \(23 - 2(3 - x)^2\)
SC B2 for \(-2(x - 3)^2\) + constant or \(23 - 2(x + \text{constant})^2\)
SC B1 for \(-2(x + 3)^2\) +constant
| Scheme | Marks |
|---|---|
| \(a + b(x^2 + 2cx + c^2)\) | M1 |
| \(2bc = 12\) or \(a + bc^2 = 5\) or \(b = -2\) | M1 |
| \(2 \times -2 \times c = 12\) or \(c = -3\) | M1 |
| \(a + -2 \times (-3)^2 = 5\) or \(a = 23\) seen Answer: \(23 - 2(x - 3)^2\) | M1 |
Notes
M1: Multiplying out expression correctly
M1: Equating coefficients or stating value of \(b\)
M1: Method to calculate \(c\)
M1: Method to calculate \(a\)
SC B3 for \(23 - 2(3 - x)^2\)