## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega = \dfrac{z}{2}+1 \Rightarrow z = 2(\omega - 1)$ | B1 | Substitution |
| $(2(\omega-1))^3 - 5(2(\omega-1))^2 + 3(2(\omega-1)) - 4 = 0$ | M1 | Substitute their expression for $z$ into cubic and attempt to expand |
| $\Rightarrow 4\omega^3 - 22\omega^2 + 35\omega - 19 = 0$ | A4 | Minus 1 each error (allow integer multiples) |
| **[6]** | | |

**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha+\beta+\gamma = 5$, $\alpha\beta+\alpha\gamma+\beta\gamma = 3$, $\alpha\beta\gamma = 4$ | B1 | Correct sums and products of roots |
| $k+l+m = \tfrac{1}{2}(\alpha+\beta+\gamma)+3 = \tfrac{11}{2} = \dfrac{-B}{A}$ | M1 | Attempt to use root relations of original equation to find all three sums and products of roots in related equation |
| $kl+km+lm = \tfrac{1}{4}(\alpha\beta+\alpha\gamma+\beta\gamma) + (\alpha+\beta+\gamma)+3 = \tfrac{35}{4} = \dfrac{C}{A}$ | | |
| $klm = \tfrac{1}{8}\alpha\beta\gamma + \tfrac{1}{4}(\alpha\beta+\beta\gamma+\beta\gamma) + \tfrac{1}{2}(\alpha+\beta+\gamma)+1 = \tfrac{19}{4} = \dfrac{-D}{A}$ | | |
| $\Rightarrow \omega^3 - \tfrac{11}{2}\omega^2 + \tfrac{35}{4}\omega - \tfrac{19}{4} = 0$ | | (*) |
| $\Rightarrow 4\omega^3 - 22\omega^2 + 35\omega - 19 = 0$ | A4 | SC (*) A3; Minus 1 each error (allow integer multiples) |
| **[6]** | | |

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