| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Express as product with specific form |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard AS-level techniques: factor theorem verification (routine substitution), algebraic factorization with a given form, and interpreting a graph to solve inequalities. All parts follow predictable methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(\frac{3}{2}) = 0\), so \((3 - 2x)\) is a factor | M1, A1 | Substituting \(\frac{3}{2}\) into \(f(x)\); Requires a correct statement and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (3 - 2x)(x + 1)^2\) | M1, A1, M1, A1 | Attempts to divide \(f(x)\) by \((3 - 2x)\) and expect to see \((3 - 2x)(x^2 + \cdots \pm 1)\); Correct quadratic factor is \((x^2 + 2x + 1)\); Attempts to factorise their \(x^2 + 2x + 1\); Correct answer only \((3 - 2x)(x + 1)^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| i. \(f(x) \leq 0\) at \(x \geq \frac{3}{2}\) and \(x = -1\) | M1, A1 | One correct answer; Both correct answers |
| ii. Roots are at \(x = -2, x = 3\) | B1 | Both correct |
### Part a:
$f(\frac{3}{2}) = 0$, so $(3 - 2x)$ is a factor | M1, A1 | Substituting $\frac{3}{2}$ into $f(x)$; Requires a correct statement and conclusion
### Part b:
$f(x) = (3 - 2x)(x + 1)^2$ | M1, A1, M1, A1 | Attempts to divide $f(x)$ by $(3 - 2x)$ and expect to see $(3 - 2x)(x^2 + \cdots \pm 1)$; Correct quadratic factor is $(x^2 + 2x + 1)$; Attempts to factorise their $x^2 + 2x + 1$; Correct answer only $(3 - 2x)(x + 1)^2$
### Part c:
**i.** $f(x) \leq 0$ at $x \geq \frac{3}{2}$ and $x = -1$ | M1, A1 | One correct answer; Both correct answers
**ii.** Roots are at $x = -2, x = 3$ | B1 | Both correct$f(x) = -2x^3 - x^2 + 4x + 3$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $(3 - 2x)$ is a factor of $f(x)$. [2]
\item Hence show that $f(x)$ can be written in the form $f(x) = (3 - 2x)(x + a)^2$ where $a$ is an integer to be found. [4]
\end{enumerate}
\includegraphics{figure_3}
Figure 3 shows a sketch of part of the curve with equation $y = f(x)$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use your answer to part (b), and the sketch, to deduce the values of $x$ for which
\begin{enumerate}[label=(\roman*)]
\item $f(x) \leq 0$
\item $f'(\frac{x}{2}) = 0$
\end{enumerate}
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q9 [9]}}