| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2024 |
| Session | October |
| Marks | 9 |
| Topic | Factor & Remainder Theorem |
| Type | Show equation reduces to polynomial |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question combining polynomial factorization with logarithm manipulation. Part (i) involves routine factor theorem verification and solving a quadratic (standard A-level techniques). Part (ii) requires applying log laws to transform the equation into the given polynomial form, then using domain restrictions to identify the valid root. While it spans multiple topics and requires careful algebraic manipulation, each step follows standard procedures without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
In this question you must show detailed reasoning.
The polynomial $f(x)$ is given by
$$f(x) = x^3 + 6x^2 + x - 4.$$
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Show that $(x + 1)$ is a factor of $f(x)$.
[1]
\item Hence find the exact roots of the equation $f(x) = 0$.
[4]
\end{enumerate}
\item \begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$2\log_2(x + 3) + \log_2 x - \log_2(4x + 2) = 1$$
can be written in the form $f(x) = 0$.
[3]
\item Explain why the equation
$$2\log_2(x + 3) + \log_2 x - \log_2(4x + 2) = 1$$
has only one real root and state the exact value of this root.
[1]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2024 Q5 [9]}}