AS June 2019 Paper 1 Q2
2. The cubic equation
\[2x^3 + 6x^2 - 3x + 12 = 0\]has roots \(\alpha\), \(\beta\) and \(\gamma\).
Without solving the equation, find the cubic equation whose roots are \((\alpha + 3)\), \((\beta + 3)\) and \((\gamma + 3)\), giving your answer in the form \(pw^3 + qw^2 + rw + s = 0\), where \(p\), \(q\), \(r\) and \(s\) are integers to be found. (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\{w = x + 3 \Rightarrow\}\ x = w - 3\) | B1 | 3.1a |
| \(2(w - 3)^3 + 6(w - 3)^2 - 3(w - 3) + 12\ (= 0)\) | M1 | 1.1b |
| \(2w^3 - 18w^2 + 54w - 54 + 6(w^2 - 6w + 9) - 3w + 9 + 12\ (= 0)\) | ||
| \(2w^3 - 12w^2 + 15w + 21 = 0\) (So \(p = 2\), \(q = -12\), \(r = 15\) and \(s = 21\)) | M1 A1 A1 | 3.1a 1.1b 1.1b |
| (5) | ||
| (5 marks) |
Notes
B1: Selects the method of making a connection between \(x\) and \(w\) by writing \(x = w - 3\)
M1: Applies the process of substituting their \(x = aw \pm b\) into \(2x^3 + 6x^2 - 3x + 12\ (= 0)\). So accept e.g. if \(x = \dfrac{w}{3}\) is used.
M1: Depends on having attempted substituting either \(x = w - 3\) or \(x = w + 3\) into the equation. This mark is for manipulating their resulting equation into the form \(pw^3 + qw^2 + rw + s\ (= 0)\) \((p \ne 0)\). The “\(= 0\)” may be implied for this.
A1: At least three of \(p\), \(q\), \(r\) and \(s\) are correct in an equation with integer coefficients. (need not have “\(= 0\)”)
A1: Correct final equation, including “\(= 0\)”. Accept integer multiples.
Alternative (ALT 1)
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \gamma = -\dfrac{6}{2} = -3,\quad \alpha\beta + \beta\gamma + \alpha\gamma = -\dfrac{3}{2},\quad \alpha\beta\gamma = -\dfrac{12}{2} = -6\) | B1 | 3.1a |
| sum roots \(= \alpha + 3 + \beta + 3 + \gamma + 3\) \(= \alpha + \beta + \gamma + 9 = -3 + 9 = 6\) pair sum \(= (\alpha + 3)(\beta + 3) + (\alpha + 3)(\gamma + 3) + (\beta + 3)(\gamma + 3)\) \(= \alpha\beta + \alpha\gamma + \beta\gamma + 6(\alpha + \beta + \gamma) + 27\) \(= -\dfrac{3}{2} + 6 \times {-3} + 27 = \dfrac{15}{2}\) product \(= (\alpha + 3)(\beta + 3)(\gamma + 3)\) \(= \alpha\beta\gamma + 3(\alpha\beta + \alpha\gamma + \beta\gamma) + 9(\alpha + \beta + \gamma) + 27\) \(= -6 + 3 \times {-\dfrac{3}{2}} + 9 \times {-3} + 27 = -\dfrac{21}{2}\) | M1 | 3.1a |
| \(w^3 - 6w^2 + \dfrac{15}{2}w - \left(-\dfrac{21}{2}\right)\ (= 0)\) | M1 | 1.1b |
| \(2w^3 - 12w^2 + 15w + 21 = 0\) (So \(p = 2\), \(q = -12\), \(r = 15\) and \(s = 21\)) | A1 A1 | 1.1b 1.1b |
| (5) |
B1: Selects the method of giving three correct equations each containing \(\alpha\), \(\beta\) and \(\gamma\).
M1: Applies the process of finding sum roots, pair sum and product.
M1: Applies \(w^3 - (\text{their sum roots})w^2 + (\text{their pair sum})w - (\text{their product})\ (= 0)\). Must be correct identities, but if quoted allow slips in substitution, but the “\(= 0\)” may be implied.
A1: At least three of \(p\), \(q\), \(r\) and \(s\) are correct in an equation with integer coefficients. (need not have “\(= 0\)”)
A1: Correct final equation, including “\(= 0\)”. Accept multiples with integer coefficients.
Note: may use another variable than \(w\) for the first four marks, but the final equation must be in terms of \(w\)
Notes: Do not isw the final two A marks – if subsequent division by 2 occurs then mark the final answer.