June 2025 Paper 1 Q4
4. Given that
- \(\mathrm{f}(x) = 2x^3 + 3x^2 - 16x + 16\)
- \(\mathrm{f}(-4) = 0\)
| Scheme | Marks | AO |
|---|---|---|
| States or uses \(a = 4\) | B1 | 1.1a |
| Valid method to find \(Q(x)\) | M1 | 2.1 |
| \((x + 4)(2x^2 - 5x + 4)\) | A1 | 1.1b |
| (3) |
Notes
B1: States or uses \(a = 4\) e.g. may be seen in their attempt at dividing algebraically by \(x + 4\)
M1: Attempts to divide \(\mathrm{f}(x)\) by \((x + 4)\) to find a three-term quadratic \(Q(x)\). There are various methods or ways to present their solution so typically methods
- by inspection look for \(2x^3 + 3x^2 - 16x + 16 = (x + 4)(2x^2 + \ldots x \pm 4)\)
- by division look for a quadratic quotient of \(2x^2 - 5x \pm \ldots\)
A1: \((x + 4)(2x^2 - 5x + 4)\) condone the missing trailing bracket i.e. \((x + 4)(2x^2 - 5x + 4\) isw if they attempt to factorise their quadratic factor.
Allow to be scored if seen in (b).
| Scheme | Marks | AO |
|---|---|---|
| Attempts to show that their \(2x^2 - 5x + 4\) does not have any (real) roots | M1 | 3.1a |
| Correct calculations, reason and conclusion | A1 | 2.1 |
| (2) | ||
| (5 marks) |
Notes
M1: Attempts to show that their three-term quadratic “\(2x^2 - 5x + 4\)” does not have any roots:
- Attempts the discriminant
e.g. \(b^2 - 4ac = 25 - 4 \times 2 \times 4\ (= -7)\) (may be embedded in the quadratic formula) - Attempts to use the quadratic formula
e.g. \((x =)\ \dfrac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times 4}}{4}\) but do not allow directly from a calculator \(\dfrac{5 \pm \sqrt{7}i}{4}\) - Attempts to complete the square
e.g. \(2x^2 - 5x + 4 = 2\left(x^2 - \dfrac{5}{2}x\right) + 4 = 2\left(x - \dfrac{5}{4}\right)^2 + \ldots\ \left(= 2\left(\left(x - \dfrac{5}{4}\right)^2 - \dfrac{25}{8} + 4\right)\right)\) - Uses calculus to find the turning point
e.g. \(\dfrac{\mathrm{d}(2x^2 - 5x + 4)}{\mathrm{d}x} = 4x - 5 = 0 \Rightarrow x = \dfrac{5}{4} \Rightarrow y = \ldots\)
Note that any attempts using the discriminant or quadratic formula must have the values embedded in the correct places (may be partially evaluated) to score M1
A1: Dependent on a correct \(Q(x) = 2x^2 - 5x + 4\)
Fully correct argument that requires:
- Fully correct work
- A justification depending on strategy and no incorrect reasoning seen
- A conclusion
Examples below – Note we must see working before they proceed to a correct root or minimum value – see M1 for guidance
| Strategy | Correct work examples | Justification examples | Conclusion examples |
|---|---|---|---|
| Via discriminant | \(b^2 - 4ac = -7\) | \(-7 \lt 0\) \(-7\) so no (real) roots but NOT \(-7 \neq 0\) so no roots | so “\(-4\) is the only (real) root” / “only one (real) root” |
| Via using the quadratic formula | \(x = \dfrac{5 \pm \sqrt{7}i}{4}\) or \(x = \dfrac{5 \pm \sqrt{-7}}{4}\) | \(-7 \lt 0\) / which is not possible / complex roots o.e. / cannot square root a negative / no (real) roots | |
| Via completing the square | \(2\left(x - \dfrac{5}{4}\right)^2 + \dfrac{7}{8}\) or \(2\left(x - \dfrac{5}{4}\right)^2 + \dfrac{7}{8} = 0\) \(\Rightarrow\) \(2\left(x - \dfrac{5}{4}\right)^2 = -\dfrac{7}{8}\) | which has a minimum value of \(\dfrac{7}{8}\) / minimum (of the positive quadratic) is above the \(x\)-axis \(-\dfrac{7}{8} \lt 0\) / cannot square root a negative / no (real) roots | |
| Via calculus | \(x = \dfrac{5}{4} \Rightarrow y = \dfrac{7}{8}\) | which has a minimum value of \(\dfrac{7}{8}\) / minimum (of the positive quadratic) is above the \(x\)-axis |
Note that it is possible to justify and conclude in one step by using phrases e.g. “no more (real) roots” or “no other (real) roots”
e.g. \(2x^2 - 5x + 4 \Rightarrow b^2 - 4ac = 25 - 32 = -7\) so no more roots which scores M1A1
Condone \(25 - 32 \lt 0\) as a justification that the quadratic has no real roots
Condone the conclusion \(-4\) is the only (real) solution (instead of (real) root).
Note if \((x + 4)\) is described as a root or \(-4\) is described as a factor this scores A0