| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining routine factor theorem application (part a), algebraic factorization (part b), and a standard trigonometric substitution (part c). While part (c) requires recognizing the substitution sin θ = x and solving within a given domain, all steps follow predictable patterns with no novel insight required. The question is slightly easier than average due to its highly scaffolded structure. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(\pm2)=0 \Rightarrow 2(\pm2)^3 + a(\pm2)^2 + 18(\pm2) - 8 = 0\) | M1 | Attempts \(g(\pm2)=0\); alternatively divides by \((x+2)\) and sets remainder equal to 0 |
| \(\Rightarrow 4a = -12 \Rightarrow a = -3\) | A1* | \(a=-3\) or equivalent following correct linear equation in \(4a\); accept \(4a=-12\) or \(4a+12=0\); this is a given answer so candidate must proceed from \(-16+4a+36-8=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(x) = 2x^3-3x^2-18x-8 = (x+2)(2x^2-7x-4)\) | M1, A1 | M1: attempts to divide \(g(x)\) by \((x+2)\); A1: correct quadratic factor \((2x^2-7x-4)\) |
| \(= (x+2)(2x+1)(x-4)\) | M1, A1 | M1: attempts to factorise quadratic; A1: \(g(x)=(x+2)(2x+1)(x-4)\); accept \(2(x+2)\left(x+\frac{1}{2}\right)(x-4)\); all factors must appear on same line; candidates writing roots without algebra score 0000 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\theta = -\frac{1}{2}\) only | B1ft | States/implies \(\sin\theta = -\frac{1}{2}\) only; follow through on all roots where \(-1\leq\sin\theta\leq1\); as long as they don't find values from \(\sin\theta=4\) or \(\sin\theta=-2\) implying they "chose" \(\sin\theta=-\frac{1}{2}\) |
| \(\theta = \frac{7}{6}\pi, \frac{11}{6}\pi\) | M1, A1 | M1: correct method for \(\sin\theta=k\), \(-1\leq k\leq1\) by arcsin (implied by \(\theta=-30°\)); A1: both values correct and exact; ignore answers outside range; condone \(1.1\dot{6}\) for \(\frac{7}{6}\) and \(1.8\dot{3}\) for \(\frac{11}{6}\) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(\pm2)=0 \Rightarrow 2(\pm2)^3 + a(\pm2)^2 + 18(\pm2) - 8 = 0$ | M1 | Attempts $g(\pm2)=0$; alternatively divides by $(x+2)$ and sets remainder equal to 0 |
| $\Rightarrow 4a = -12 \Rightarrow a = -3$ | A1* | $a=-3$ or equivalent following correct linear equation in $4a$; accept $4a=-12$ or $4a+12=0$; this is a given answer so candidate **must** proceed from $-16+4a+36-8=0$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(x) = 2x^3-3x^2-18x-8 = (x+2)(2x^2-7x-4)$ | M1, A1 | M1: attempts to divide $g(x)$ by $(x+2)$; A1: correct quadratic factor $(2x^2-7x-4)$ |
| $= (x+2)(2x+1)(x-4)$ | M1, A1 | M1: attempts to factorise quadratic; A1: $g(x)=(x+2)(2x+1)(x-4)$; accept $2(x+2)\left(x+\frac{1}{2}\right)(x-4)$; all factors must appear on same line; candidates writing roots without algebra score 0000 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = -\frac{1}{2}$ only | B1ft | States/implies $\sin\theta = -\frac{1}{2}$ only; follow through on all roots where $-1\leq\sin\theta\leq1$; as long as they don't find values from $\sin\theta=4$ or $\sin\theta=-2$ implying they "chose" $\sin\theta=-\frac{1}{2}$ |
| $\theta = \frac{7}{6}\pi, \frac{11}{6}\pi$ | M1, A1 | M1: correct method for $\sin\theta=k$, $-1\leq k\leq1$ by arcsin (implied by $\theta=-30°$); A1: both values correct and exact; ignore answers outside range; condone $1.1\dot{6}$ for $\frac{7}{6}$ and $1.8\dot{3}$ for $\frac{11}{6}$ |
---7.
$$g ( x ) = 2 x ^ { 3 } + a x ^ { 2 } - 18 x - 8$$
Given that $( x + 2 )$ is a factor of $\mathrm { g } ( x )$,
\begin{enumerate}[label=(\alph*)]
\item show that $a = - 3$
\item Hence, using algebra, fully factorise $\mathrm { g } ( x )$.
Using your answer to part (b),
\item solve, for $0 \leqslant \theta < 2 \pi$, the equation
$$2 \sin ^ { 3 } \theta - 3 \sin ^ { 2 } \theta - 18 \sin \theta = 8$$
giving each answer, in radians, as a multiple of $\pi$.\\
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\hfill \mbox{\textit{Edexcel C12 2017 Q7 [9]}}