| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Show equation reduces to polynomial |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard techniques: (a) routine Factor Theorem application requiring substitution of x=3/4, (b) algebraic manipulation using double angle formulas (standard C3/C4 content), and (c) combining parts (a) and (b) with consideration of the range of cosine. While it requires multiple techniques, each step follows predictable patterns with clear signposting, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
6
\begin{enumerate}[label=(\alph*)]
\item Use the Factor Theorem to show that $4 x - 3$ is a factor of
$$16 x ^ { 3 } + 11 x - 15$$
\item Given that $x = \cos \theta$, show that the equation
$$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$
can be written in the form
$$16 x ^ { 3 } + 11 x - 15 = 0$$
\item Hence show that the only solutions of the equation
$$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$
are given by $\cos \theta = \frac { 3 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2012 Q6 [10]}}