June 2025 Paper 1 Q19
19
(a)
\((x - 1)\) is a factor of \(\quad 3x^3 + 2x^2 - 3x - 2\)
Factorise fully \(\quad 3x^3 + 2x^2 - 3x - 2\) [3 marks]
(b)
\(\mathrm{f}(x) = (x - 2)(5x + 4)(3x + 2)\)
\(c\) is a positive integer.
The value of \(\mathrm{f}(c)\) can be written as the product of two prime factors.
Deduce the value of \(\mathrm{f}(c)\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: inspection or long division | ||
| \(3x^2 + kx + 2\) by inspection or first two terms using long division: \(3x^2 + 5x + k\) | M1 | \(k\) could be blank. Can be seen in a grid method |
| \(3x^2 + 5x + 2\) | M1dep | all terms correct written as a single expression |
| \((x - 1)(3x + 2)(x + 1)\) | A1 | any order of brackets |
| Alternative method 2: factor theorem | ||
| Any one of \(\mathrm{f}(-1)\), \(\mathrm{f}(2)\), \(\mathrm{f}(-2)\), \(\mathrm{f}\left(-\dfrac{2}{3}\right)\) attempted and evaluated correctly | M1 | \(\mathrm{f}(-1) = 0\), \(\mathrm{f}(2) = 24\), \(\mathrm{f}(-2) = -12\), \(\mathrm{f}\left(-\dfrac{2}{3}\right) = 0\) |
| \((3x + 2)\) or \((x + 1)\) identified as a factor | M1dep | oe |
| \((x - 1)(3x + 2)(x + 1)\) | A1 | any order of brackets |
| Alternative method 3: equating coefficients | ||
| \((x - 1)(ax^2 + bx + c)\) and \(ax^3 - ax^2 + bx^2 - bx + cx - c\) | M1 | |
| \(a = 3\), \(b = 5\) and \(c = 2\) or \(3x^2 + 5x + 2\) | M1dep | |
| \((x - 1)(3x + 2)(x + 1)\) | A1 | any order of brackets |
| Alternative method 4: factorising without using \((x - 1)\) | ||
| \(x^2(3x + 2) - 1(3x + 2)\) | M1 | 1 could be omitted |
| \((3x + 2)(x^2 - 1)\) | M1dep | |
| \((x - 1)(3x + 2)(x + 1)\) | A1 | any order of brackets |
Additional guidance
| Penalise further working to find solutions to \(\mathrm{f}(x) = 0\) | M2A0 |
| Correct answer with no working | M2A1 |
| \((x - 1)\) or \((3x + 2)\) or \((x + 1)\) on answer line on their own with no working | M0A0 |
| \((3x + 2)(x + 1)\) on its own on answer line | M2A0 |
| Condone missing brackets recovered | |
| Do not award A mark for missing brackets not recovered | |
| In Alt 1 synthetic division can be used |
| Answer | Mark | Comments |
|---|---|---|
| \(x\) or \(c = 3\) used | M1 | eg \(1 \times 19 \times 11\) or 19 and 11 seen together in working |
| 209 or \(19 \times 11\) | A1 |
Additional guidance
| \(c = 209\) would imply method | M1A0 |
| \(x\) or \(c = 3\) could be one of a number of trials | M1 |
| \(19 \times 11\) seen in working but then worked out incorrectly for answer | M1A0 |
| \((c =)\ 3\) on answer line not from incorrect working | M1A0 |