# Question 10:

## Question 10(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = 3z - 1 \Rightarrow z = \dfrac{w+1}{3}$ | B1 | 3.1a — Selects the method of making a connection between $z$ and $w$ by writing $z = \dfrac{w+1}{3}$. Other variables may be used |
| $\left(\dfrac{w+1}{3}\right)^4 + 5\left(\dfrac{w+1}{3}\right)^2 - 30 = 0$ | M1 | 3.1a — Applies the process of substituting their $z = \dfrac{w+1}{3}$ into $z^4 + 5z^2 - 30 = 0$ |
| $\dfrac{1}{81}(w^4 + 4w^3 + 6w^2 + 4w + 1) + \dfrac{5}{9}(w^2 + 2w + 1) - 30 = 0$ leading to $w^4 + aw^3 + bw^2 + cw + d\ \{= 0\}$ | M1 | 1.1b — Manipulates equation into the form $w^4 + aw^3 + bw^2 + cw + d = 0$. Note "= 0" can be missing |
| $w^4 + 4w^3 + 51w^2 + 94w - 2384 = 0$ | A1 A1 | 1.1b — First A1: at least two of $a, b, c, d$ correct ("= 0" can be missing). Second A1: fully correct equation including "= 0", must be in terms of $w$ |
| **(5 marks)** | | |

**Alternative Method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $p+q+r+s=0,\quad pq+pr+ps+qr+qs+rs=5$ $pqr+pqs+prs+qrs=0,\quad pqrs=-30$ | B1 | 3.1a — Gives four correct equations containing $p, q, r, s$ |
| New sum $= 3(p+q+r+s)-4 = \ldots\{-4\}$ New pair sum $= 9(pq+pr+\ldots+rs)-9(p+q+r+s)+6=\ldots\{51\}$ New triple sum $= 27(pqr+pqs+prs+qrs)-18(pq+pr+ps+qr+qs+rs)+6(p+q+r+s)-4=\ldots\{-94\}$ New product $= 81(pqrs)-27(pqr+pqs+prs+qrs)+9(pq+pr+ps+qr+qs+rs)-3(p+q+r+s)+1 = \ldots\{-2384\}$ | M1 | 3.1a — Applies the process of finding **at least three** of the new sum, new pair sum, new triple sum and new product |
| Applies $w^4 - (\text{new sum})w^3 + (\text{new pair sum})w^2 - (\text{new triple sum})w + (\text{new product}) = 0$ | M1 | 1.1b — Note "= 0" can be missing |
| $w^4 + 4w^3 + 51w^2 + 94w - 2384 = 0$ | A1 A1 | 1.1b — As above |
| **(5 marks)** | | |

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## Question 10(ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha + 2\alpha + \alpha - \beta = 0$ and $\alpha \times 2\alpha \times (\alpha - \beta) = -\dfrac{81}{4}$ | M1 A1 | 3.1a, 1.1b — Uses the sum and product to form two equations in $\alpha$ and $\beta$. Condone product $= \dfrac{81}{4}$ for M1. Note: $4\alpha - \beta = -\dfrac{n}{4}$ or $4\alpha - \beta = 81$ is M0. A1: correct equations need not be simplified |
| Solves simultaneously e.g. $4\alpha - \beta = 0 \Rightarrow \beta = 4\alpha$ $2\alpha^2(\alpha - 4\alpha) = -\dfrac{81}{4} \Rightarrow \alpha^3 = \dfrac{27}{8} \Rightarrow \alpha = \ldots$ | M1 | 3.1a — Solves simultaneous equations to find a value for $\alpha$ or $\beta$ |
| Uses their values $\alpha = \dfrac{3}{2},\ \beta = 6$ to find the roots $\alpha, 2\alpha, \alpha - \beta$ | M1 | 1.1b — Condone third root as $\beta$ |
| Roots $1.5,\ 3,\ -4.5$ | A1 | 1.1b — Correct roots |
| **(5 marks)** | | |

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## Question 10(ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n = [(1.5 \times 3) + (1.5 \times -4.5) + (3 \times -4.5)] \times 4$ **Or** multiplies out $(x-3)\!\left(x-\dfrac{3}{2}\right)\!\left(x+\dfrac{9}{2}\right)$ or $(x-3)(2x-3)(2x+9)$ to achieve the form $4x^3 + \ldots$ | M1 | 1.1b — Finds the pair sum for their numerical roots and multiplies by 4. Alternative: multiplies out three brackets $(x - \text{their } \alpha)(x - \text{their } 2\alpha)(x-(\alpha-\beta))$ to achieve the form $4x^3 + \ldots$ |
| $n = -63$ (cso, must have correct roots in (a)) | A1 | 1.1b — Correct value **from correct roots only** |
| **(2 marks)** | | |
| **Total: 12 marks** | | |