Specimen 2020 Paper 2 Q19
19
\[\mathrm{f}(x) = 3x^3 - 2x^2 - 7x - 2\](a) Use the factor theorem to show that \(\ (3x + 1)\ \) is a factor of \(\mathrm{f}(x)\). [2 marks]
(b) Factorise \(\mathrm{f}(x)\) fully. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Identifies \((x =)\ -\dfrac{1}{3}\) | M1 | may be implied |
| \(3\left(-\dfrac{1}{3}\right)^3 - 2\left(-\dfrac{1}{3}\right)^2 - 7\left(-\dfrac{1}{3}\right) - 2 = 0\) or \(-\dfrac{1}{9} - \dfrac{2}{9} + \dfrac{7}{3} - 2 = 0\) | A1 | oe must show four terms and equate to 0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((3x + 1)(x^2 - x \ldots)\) or \(\begin{array}{l} \phantom{3x + 1\,)\,}x^2 - x \ldots \\ 3x + 1\,\overline{)\,3x^3 - 2x^2 - 7x - 2} \end{array}\) | M1 | |
| \((3x + 1)(x^2 - x - 2)\) or \(\begin{array}{l} \phantom{3x + 1\,)\,}x^2 - x - 2 \\ 3x + 1\,\overline{)\,3x^3 - 2x^2 - 7x - 2} \end{array}\) | A1 | |
| \((3x + 1)(x + 1)(x - 2)\) | A1 | |
| Alternative method 2 | ||
| \(\mathrm{f}(-1) = 0\) or \(\mathrm{f}(2) = 0\) | M1 | |
| \(\mathrm{f}(-1) = 0\) and \(\mathrm{f}(2) = 0\) | A1 | |
| \((3x + 1)(x + 1)(x - 2)\) | A1 | |
Notes
In Alternative method 1 the long division is printed with \(3x^3 + 4x^2 - 2x - 1\) under the division sign (corrected from the printed mark scheme: the cubic being divided is \(\mathrm{f}(x) = 3x^3 - 2x^2 - 7x - 2\)).