# Question 3:

## Method 1 (Substitution):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\omega = 2x-1 \Rightarrow x = \frac{\omega+1}{2}$ | M1 | Using a substitution |
| Correct substitution | A1 | Correct |
| $\left(\frac{\omega+1}{2}\right)^3 - 4\left(\frac{\omega+1}{2}\right)^2 + 8\left(\frac{\omega+1}{2}\right) + 3 = 0$ | M1 | Substitute into cubic |
| $\Rightarrow \frac{1}{8}(\omega^3+3\omega^2+3\omega+1)-(\omega^2+2\omega+1)+4(\omega+1)+3=0$ | M1 | Attempting to expand cubic and quadratic |
| $\Rightarrow \omega^3 - 5\omega^2 + 19\omega + 49 = 0$ | A2, A1 [7] | LHS oe, -1 each error; Correct equation |

## Method 2 (Root relationships):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha+\beta+\gamma=4$, $\alpha\beta+\alpha\gamma+\beta\gamma=8$, $\alpha\beta\gamma=-3$ | M1, A1 | Attempt to find $\Sigma\alpha$, $\Sigma\alpha\beta$, $\alpha\beta\gamma$; All correct |
| $k+l+m = 2(\alpha+\beta+\gamma)-3 = 5 = \frac{-B}{A}$ | M1 | Attempt to use root relationships to find at least two of $\Sigma k$, $\Sigma kl$, $klm$ |
| $kl+km+lm = 4(\alpha\beta+\alpha\gamma+\beta\gamma)-4(\alpha+\beta+\gamma)+3 = 19 = \frac{C}{A}$ | | |
| $klm = 8\alpha\beta\gamma - 4(\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)-1 = -49 = \frac{-D}{A}$ | | |
| $\Rightarrow \omega^3 - 5\omega^2 + 19\omega + 49 = 0$ | A1, A1, A1, A1 [7] | Quadratic coefficient; Linear coefficient; Constant term; Correct equation |

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