Higher June 2023 Paper 5 Q15
15
(a) Factorise.\[9x^2 - 4\] [2]
(b) Solve by factorisation.\[3x^2 - 2x - 8 = 0\] [3]
(c) Solve.\[\frac{2(x - 5)}{1 - 3x} = 2\] [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((3x + 2)(3x - 2)\) final answer | 2 | M1 for answer a pair of factors of the type \((ax + b)(ax - b)\), where \(a = 3\) or \(b = 2\) or for correct answer seen | For 2 marks or M1, condone omission of final bracket |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((3x + 4)(x - 2)\) | M2 | M1 for \(3x(x - 2) + 4(x - 2)\) or \(x(3x + 4) - 2(3x + 4)\) or for \((3x + a)(x + b)\) where \(ab = -8\) or \(3b + a = -2\) | For M2 and M1 condone omission of final bracket For M2, condone \(\frac{(3x - 6)(3x + 4)}{3}\) followed by \(x - 2\) [= 0] and \(3x + 4\) [= 0] If no product of factors shown, \(x - 2 = 0\) and \(3x + 4 = 0\) gets M1 only After \(3x(x - 2) + 4(x - 2)\) or \(x(3x + 4) - 2(3x + 4)\) and then correct answers allow M2B1 |
| \(-\frac{4}{3}\) oe and 2 | B1 | Correct or FT their two factors dep on factors of the form \((3x \pm a)(x \pm b)\) | Accept –1.33… |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.5 or \(1\frac{1}{2}\) or \(\frac{3}{2}\) | 4 | M1 for \(2(x - 5) = 2(1 - 3x)\) or \(\frac{x - 5}{1 - 3x} = 1\) M1 for \(2x - 10 = 2 - 6x\) or \(x - 5 = 1 - 3x\) M1 for reaching \(ax = b\), FT their previous working provided previous working is of the form \(dx + e = f + gx\) | For 4 marks, condone \(\frac{12}{8}\) or \(\frac{6}{4}\) isw incorrect cancelling/conversion Embedded answer scores M3 maximum This final method mark may be implied from the answer |