# Question 6 (Method 1):

| $w = 2x \Rightarrow x = \frac{w}{2}$ | B1 | Substitution. For $x = 2w$ give B0 but follow through for max 3 marks |
| $\Rightarrow 2\left(\frac{w}{2}\right)^3 + \left(\frac{w}{2}\right)^2 - 3\left(\frac{w}{2}\right) + 1 = 0$ | M1, A1 | Substitute into cubic; Correct substitution |
| $\Rightarrow w^3 + w^2 - 6w + 4 = 0$ | A2 [5] | Minus 1 for each error (including $= 0$ missing), minimum 0; Give full credit for integer multiple of equation |

## Question 6 (OR Method):

| $\alpha + \beta + \gamma = -\frac{1}{2}$, $\alpha\beta + \alpha\gamma + \beta\gamma = -\frac{3}{2}$, $\alpha\beta\gamma = -\frac{1}{2}$ | B1 | All three |
| Let new roots be $k, l, m$: $k+l+m = 2(\alpha+\beta+\gamma) = -1 = \frac{-B}{A}$ | M1 | Attempt to use sums and products of roots of original equation to find sums and products of roots in related equation |
| $kl+km+lm = 4(\alpha\beta+\alpha\gamma+\beta\gamma) = -6 = \frac{C}{A}$ | A1 | Sums and products all correct |
| $klm = 8\alpha\beta\gamma = -4 = \frac{-D}{A}$ | | |
| $\Rightarrow \omega^3 + \omega^2 - 6\omega + 4 = 0$ | A2 [5] | ft their coefficients; minus one for each error (including $= 0$ missing), minimum 0; Give full credit for integer multiple of equation |

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