June 2024 Paper 1 Q11
11
(a)
\((3x - 7)\) is a factor of \(\quad 3x^3 - 4x^2 - 13x + 14\)
Work out the other two linear factors. [3 marks]
(b)
\((x - 2)\) is a factor of \(\quad ax^4 - 3ax^3 + 5x - 22\)
Work out the value of \(a\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – by inspection or long division | ||
| \(x^2 + kx - 2\) by inspection or first two terms using long division: \(x^2 + x + k\) | M1 | \(k\) could be blank can be seen in a grid method |
| \(x^2 + x - 2\) | A1 | |
| \((x - 1)\) and \((x + 2)\) or \((x - 1)(x + 2)\) or \((3x - 7)(x - 1)(x + 2)\) | A1 | any order of brackets |
| Alternative method 2 – factor theorem | ||
| 1 or 2 or \(-1\) or \(-2\) substituted into function and evaluated correctly | M1 | \(\mathrm{f}(x) = 0\) for \(x = 1\) \(\mathrm{f}(x) = 0\) for \(x = -2\) \(\mathrm{f}(x) = 20\) for \(x = -1\) \(\mathrm{f}(x) = -4\) for \(x = 2\) ignore additional incorrect calculations |
| \((x + 2)\) or \((x - 1)\) identified as a factor | A1 | must have come from a substitution from M mark |
| \((x - 1)\) and \((x + 2)\) or \((x - 1)(x + 2)\) or \((3x - 7)(x - 1)(x + 2)\) | A1 | any order of brackets |
| Alternative method 3 – equating coefficients | ||
| \((3x - 7)(ax^2 + bx + c)\) and \(3ax^3 - 7ax^2 + 3bx^2 - 7bx + 3cx - 7c\) | M1 | |
| \(a = 1\), \(b = 1\) and \(c = -2\) or \(x^2 + x - 2\) | A1 | |
| \((x - 1)\) and \((x + 2)\) or \((x - 1)(x + 2)\) or \((3x - 7)(x - 1)(x + 2)\) | A1 | any order of brackets |
Additional guidance
| \(x = 1\) and \(-2\) on answer line or last line of working | SC2 |
| Follow whichever scheme gives the best mark | |
| Penalise incorrect further working eg. correct answer followed by \(x = 1\) and \(x = -2\) | M1A1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – by substitution | ||
| \(a \times 2^4 - 3a \times 2^3 + 5 \times 2 - 22\) or \(16a - 24a + 10 - 22\) | M1 | oe 4 terms with at least 3 correct |
| \(16a - 24a + 10 - 22 = 0\) or \(-8a = 12\) | M1dep | oe could imply first M mark |
| \(a = -1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\) | A1 | |
| Alternative method 2 – long division | ||
| First two terms correct | M1 | \(ax^3 - ax^2\) |
| Correct equation set up to find \(a\) | M1dep | eg \(-8a = 12\) or \(5 - 4a = 11\) |
| \(a = -1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\) | A1 | |
| Alternative method 3 – equating coefficients or inspection | ||
| \(d = 11\) and \(ax^3\) and \(c = 3\) or \(b = 1.5\) or \(c = -2a\) or \(b = -a\) | M1 | could be seen in a grid for \(ax^3 + bx^2 + cx + d\) |
| Correct equation set up to find \(a\) | M1dep | eg \(4a + 11 = 5\) or \(2a + 3 = 0\) |
| \(a = -1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\) | A1 | |
Additional guidance
| Follow whichever scheme gives the best mark |