June 2022 Paper 2 Q20
20 \(\mathrm{f}(x) = 2x^3 + 11x^2 + 12x - 9\)
(a) Use the factor theorem to show that \(\quad (2x - 1) \quad\) is a factor of \(\mathrm{f}(x)\). [2 marks]
(b) Show that \(\quad \mathrm{f}(x) = 0 \quad\) has exactly two solutions. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Shows substitution of \(x = \dfrac{1}{2}\) | M1 | eg \(2 \times \left(\dfrac{1}{2}\right)^3 + 11 \times \left(\dfrac{1}{2}\right)^2 + 12 \times \dfrac{1}{2} - 9\) or \(2 \times \dfrac{1}{8} + 11 \times \dfrac{1}{4} + 12 \times \dfrac{1}{2} - 9\) or \(\dfrac{1}{4} + \dfrac{11}{4} + 6 - 9\) |
| Shows substitution of \(x = \dfrac{1}{2}\) and evaluates to zero | A1 | eg \(2 \times \left(\dfrac{1}{2}\right)^3 + 11 \times \left(\dfrac{1}{2}\right)^2 + 12 \times \dfrac{1}{2} - 9 = 0\) or \(2 \times \dfrac{1}{8} + 11 \times \dfrac{1}{4} + 12 \times \dfrac{1}{2} - 9 = 0\) or \(\dfrac{1}{4} + \dfrac{11}{4} + 6 - 9 = 0\) |
Additional guidance
| Allow use of 0.5 and/or absence of multiplication signs eg1 \(\;2(0.5)^3 + 11(0.5)^2 + 12(0.5) - 9 = 0\) eg2 \(\;2\left(\dfrac{1}{8}\right) + 11\left(\dfrac{1}{4}\right) + 12\left(\dfrac{1}{2}\right) - 9\) | M1A1 M1A0 | |||||||||||||||
| Allow working in stages eg \(\;2(0.5)^3 + 11(0.5)^2 + 12(0.5) = 9 \qquad 9 - 9 = 0\) | M1A1 | |||||||||||||||
| Condone incorrect use of = eg \(\;2(0.5)^3 + 11(0.5)^2 + 12(0.5) = 9 - 9 = 0\) | M1A1 | |||||||||||||||
| Condone \(2 \times \dfrac{1}{2}^3\) or \(2 \times \left(\dfrac{1^3}{2}\right)\) etc | ||||||||||||||||
| Ignore algebraic division or other substitution attempts | ||||||||||||||||
| Only stating \(\mathrm{f}\left(\dfrac{1}{2}\right)\) or only stating \(\mathrm{f}\left(\dfrac{1}{2}\right) = 0\) | M0A0 | |||||||||||||||
| Calculation error(s) will be A0 eg1 \(\;2 \times \left(\dfrac{1}{2}\right)^3 + 11 \times \left(\dfrac{1}{2}\right)^2 + 12 \times \dfrac{1}{2} - 9 = \dfrac{1}{8} + \dfrac{11}{4} + 6 - 9 = 0\) eg2 \(\;\dfrac{1}{4} + \dfrac{11}{4} + 6 - 9 = 4 + 6 - 9 = 0\) | M1A0 M1A0 | |||||||||||||||
| May be seen as synthetic division eg
(with an error award M0A0) | M1A1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(x^2 + 6x \ldots\) or \(2 \times (-3)^3 + 11 \times (-3)^2 + 12 \times (-3)\) \(- 9\) | M1 | oe eg \(\begin{array}{r} x^2 + 6x \ldots \\ 2x - 1\,\overline{\big)\,2x^3 + 11x^2 + 12x - 9} \end{array}\) or \((2x - 1)(x^2 + bx + c)\) and \(b = 6\) or \(2 \times -27 + 11 \times 9 + 12 \times -3 - 9\) or \(-54 + 99 - 36 - 9\) |
| \(x^2 + 6x + 9\) or \((x + 3)(x + 3)\) or \((x + 3)^2\) | M1dep | oe eg \(\begin{array}{r} x^2 + 6x + 9 \\ 2x - 1\,\overline{\big)\,2x^3 + 11x^2 + 12x - 9} \end{array}\) or \((2x - 1)(x^2 + bx + c)\) and \(b = 6\) and \(c = 9\) |
| \(x^2 + 6x + 9\) and \((x + 3)(x + 3)\) or \(x^2 + 6x + 9\) and \(\dfrac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 9}}{2 \times 1}\) or \(x^2 + 6x + 9\) and \(6^2 - 4 \times 1 \times 9 = 0\) or \((2x - 1)(x + 3)(x + 3)\) | M1dep | oe eg \(x^2 + 6x + 9\) and \((x + 3)^2\) or \(x^2 + 6x + 9\) and \(\dfrac{-6}{2}\) or \(x^2 + 6x + 9\) and \(36 - 36 = 0\) or \((2x - 1)(x + 3)^2\) |
| M3 and indication that there are exactly two solutions | A1 | eg1 \(x^2 + 6x + 9\) and \((x + 3)(x + 3)\) and 0.5 and \(-3\) eg2 \(x^2 + 6x + 9\) and \(\dfrac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 9}}{2 \times 1}\) and 0.5 and \(-3\) eg3 \((2x - 1)(x + 3)(x + 3)\) and repeated bracket so exactly two solutions/roots/answers/factors |
| Alternative method 2 | ||
| \(6x^2 + 22x + 12 = 0\) or \((6x + 4)(x + 3) = 0\) or \(\dfrac{-22 \pm \sqrt{22^2 - 4 \times 6 \times 12}}{2 \times 6}\) or \(\dfrac{-22 \pm \sqrt{196}}{12}\) | M1 | condone omission of = 0 oe eg \((2x + 6)(3x + 2) = 0\) or \(2(x + 3)(3x + 2) = 0\) or \(-\dfrac{11}{6} \pm \sqrt{-2 + \dfrac{121}{36}}\) or \(-\dfrac{11}{6} \pm \sqrt{\dfrac{49}{36}}\) |
| \(x = -\dfrac{2}{3}\) and \(x = -3\) | M1dep | allow \([-0.67, -0.66]\) for \(-\dfrac{2}{3}\) |
| \(x = -\dfrac{2}{3}\) and \((-3, 0)\) | M1dep | allow \([-0.67, -0.66]\) for \(-\dfrac{2}{3}\) ignore \(y\)-coordinate for \(x = -\dfrac{2}{3}\) \((-3, 0)\) may be seen on a graph |
| M3 and indication that there are exactly two solutions | A1 | eg \(x = -\dfrac{2}{3}\) and \((-3, 0)\) and a turning point on the \(x\)-axis so two solutions/roots |
| Alternative method 3 | ||
| Sketch of cubic graph with maximum turning point at \((-3, 0)\) | M1 | condone minimum turning point at \((-3, 0)\) |
| Sketch of cubic graph with maximum turning point at \((-3, 0)\) and minimum turning point in the third quadrant | M1dep | |
| Sketch of cubic graph with maximum turning point at \((-3, 0)\) and minimum turning point in the third quadrant and intersecting the positive \(x\)-axis at \(\dfrac{1}{2}\) | M1dep | \(-3\) and \(\dfrac{1}{2}\) must both be correctly labelled on the \(x\)-axis |
| M3 and indication that there are exactly two solutions | A1 | eg M3 and 0.5 and \(-3\) |
Additional guidance
| Up to M3 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | ||||||||||||||||
| Alt 1 Up to the first two marks may be seen in a grid eg
| M1M1 | |||||||||||||||
| Alt 1 \(x^2 + 6x + 9\) or \((x + 3)(x + 3)\) or \((x + 3)^2\) | M1M1 | |||||||||||||||
| Alt 1 \((2x - 1)(x + 3)(x + 3)\) or \((2x - 1)(x + 3)^2\) | M1M1M1 | |||||||||||||||
| Alt 1 \((2x - 1)(x + 3)(x + 3)\) with solutions 0.5 and \(-3\) | M1M1M1A1 | |||||||||||||||
| Alt 1 \(2x^2 + 5x - 3 = (2x - 1)(x + 3) \qquad\) 0.5 and \(-3\) | Zero | |||||||||||||||
| Alt 1 Examples of acceptable indications that there are exactly two solutions eg1 \(\;x = 0.5, -3, -3 \quad\) (Only) two solutions eg2 \(\;x = 0.5, -3, -3 \quad\) One root is a repeat eg3 \(\;(2x - 1)\) gives one solution \(\;(x + 3)(x + 3)\) gives one solution eg4 \(\;(2x - 1)(x + 3)(x + 3) \quad\) Two factors (only) | ||||||||||||||||
| Alt 1 These are not acceptable indications that there are exactly two solutions eg1 \(\;(2x - 1)(x + 3)(x + 3) \quad\) 3 and 0.5 eg2 \(\;(x + 3)(x + 3) \quad\) Exactly two solutions | ||||||||||||||||
| Alt 1 Ignore other substitution attempts if using factor theorem for 1st M1 | ||||||||||||||||
| Alt 1 Allow absence of multiplication signs in factor theorem eg \(\;2(-3)^3 + 11(-3)^2 + 12(-3) - 9\) | M1 | |||||||||||||||
| Alt 1 Condone incorrect use of = eg \(\;2(-3)^3 + 11(-3)^2 + 12(-3) = 9 - 9\) | M1 | |||||||||||||||
| Alt 1 Allow working in stages eg \(\;2(-3)^3 + 11(-3)^2 + 12(-3) = 9 \qquad 9 - 9 = 0\) | M1 | |||||||||||||||
| Alt 1 Only stating \(\mathrm{f}(-3)\) or only stating \(\mathrm{f}(-3) = 0\) | M0 | |||||||||||||||
| Alt 1 May be seen as synthetic division eg
| M1 | |||||||||||||||
| Working in (a) eg algebraic division that is not used in (b) cannot score in (b) eg (a) \(\begin{array}{r} x^2 + 6x + 9 \\ 2x - 1\,\overline{\big)\,2x^3 + 11x^2 + 12x - 9} \end{array}\) (b) Not attempted | M0 | |||||||||||||||
| Working in (a) eg algebraic division that is used in (b) can score in (b) eg (a) \((2x - 1)(x^2 + 6x + 9)\) (b) Student shows an arrow from their working in (a) | M1M1 |