4 The roots of the cubic equation \(x ^ { 3 } - 2 x ^ { 2 } - 8 x + 11 = 0\) are \(\alpha , \beta\) and \(\gamma\).
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Question 4:
Answer Marks
Guidance
Answer Mark
Guidance
\(w = x + 1 \Rightarrow x = w - 1\) B1
Substitution. For \(x = w+1\) give B0 but then follow for a maximum of 3 marks
\(x^3 - 2x^2 - 8x + 11 = 0,\ w = x - 1\) \(\Rightarrow (w-1)^3 - 2(w-1)^2 - 8(w-1) + 11 = 0\) M1, M1
Attempt to substitute into cubic; attempt to expand
\(\Rightarrow w^3 - 5w^2 - w + 16 = 0\) A3 [6]
\(-1\) for each error (including omission of \(= 0\))
OR:
Answer Marks
Guidance
Answer Mark
Guidance
\(\alpha + \beta + \gamma = 2\); \(\alpha\beta + \alpha\gamma + \beta\gamma = -8\); \(\alpha\beta\gamma = -11\) B1
All 3 correct
Let new roots be \(k, l, m\) then: \(k+l+m = \alpha+\beta+\gamma+3 = 2+3 = 5\) M1
Valid attempt to use sum of roots in original equation to find sum of roots in new equation
\(kl+km+lm = (\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)+3 = -8+4+3 = -1\) M1
Valid attempt to use product of roots in original equation to find one of \(\sum\alpha\beta\) or \(\alpha\beta\gamma\)
\(klm = \alpha\beta\gamma + (\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha+\beta+\gamma)+1 = -11-8+2+1 = -16\) \(\Rightarrow w^3 - 5w^2 - w + 16 = 0\) A3 [6]
\(-1\) each error (including omission of \(= 0\))
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# Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $w = x + 1 \Rightarrow x = w - 1$ | B1 | Substitution. For $x = w+1$ give B0 but then follow for a maximum of 3 marks |
| $x^3 - 2x^2 - 8x + 11 = 0,\ w = x - 1$ | | |
| $\Rightarrow (w-1)^3 - 2(w-1)^2 - 8(w-1) + 11 = 0$ | M1, M1 | Attempt to substitute into cubic; attempt to expand |
| $\Rightarrow w^3 - 5w^2 - w + 16 = 0$ | A3 **[6]** | $-1$ for each error (including omission of $= 0$) |
**OR:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha + \beta + \gamma = 2$; $\alpha\beta + \alpha\gamma + \beta\gamma = -8$; $\alpha\beta\gamma = -11$ | B1 | All 3 correct |
| Let new roots be $k, l, m$ then: | | |
| $k+l+m = \alpha+\beta+\gamma+3 = 2+3 = 5$ | M1 | Valid attempt to use sum of roots in original equation to find sum of roots in new equation |
| $kl+km+lm = (\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)+3 = -8+4+3 = -1$ | M1 | Valid attempt to use product of roots in original equation to find one of $\sum\alpha\beta$ or $\alpha\beta\gamma$ |
| $klm = \alpha\beta\gamma + (\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha+\beta+\gamma)+1 = -11-8+2+1 = -16$ | | |
| $\Rightarrow w^3 - 5w^2 - w + 16 = 0$ | A3 **[6]** | $-1$ each error (including omission of $= 0$) |
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4 The roots of the cubic equation $x ^ { 3 } - 2 x ^ { 2 } - 8 x + 11 = 0$ are $\alpha , \beta$ and $\gamma$.\\
Find the cubic equation with roots $\alpha + 1 , \beta + 1$ and $\gamma + 1$.
\hfill \mbox{\textit{OCR MEI FP1 2010 Q4 [6]}}