| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Factorise polynomial completely |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on standard AS-level polynomial techniques. Part (a) requires simple substitution to verify a factor, part (b) involves routine polynomial division and factoring a quadratic, and part (c) is algebraic manipulation with partial fractions. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(-3) = (-3)^3 + 3\times(-3)^2 - 4\times(-3) - 12\) | M1 | Attempts \(f(-3)\) |
| \(f(-3) = 0 \Rightarrow (x+3)\) is a factor \(\Rightarrow f(x)\) is divisible by \((x+3)\) | A1 | Must achieve \(f(-3)=0\) and explain \((x+3)\) is a factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 + 3x^2 - 4x - 12 = (x+3)(x^2 - 4)\) | M1 | Attempts to divide by \((x+3)\) to get quadratic factor |
| \(= (x+3)(x+2)(x-2)\) | dM1, A1 | dM1 for attempting to factorise \((x^2-4)\); check first and last terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x^3+3x^2-4x-12}{x^3+5x^2+6x} = \frac{\ldots}{x(x^2+5x+6)}\) | M1 | Takes common factor \(x\) from denominator and writes numerator in factors |
| \(= \frac{(x+3)(x+2)(x-2)}{x(x+3)(x+2)}\) | dM1 | Further factorises denominator and cancels |
| \(= \frac{x-2}{x} = 1 - \frac{2}{x}\) | A1 | No errors or omissions |
## Question 5:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(-3) = (-3)^3 + 3\times(-3)^2 - 4\times(-3) - 12$ | M1 | Attempts $f(-3)$ |
| $f(-3) = 0 \Rightarrow (x+3)$ is a factor $\Rightarrow f(x)$ is divisible by $(x+3)$ | A1 | Must achieve $f(-3)=0$ and explain $(x+3)$ is a factor |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 + 3x^2 - 4x - 12 = (x+3)(x^2 - 4)$ | M1 | Attempts to divide by $(x+3)$ to get quadratic factor |
| $= (x+3)(x+2)(x-2)$ | dM1, A1 | dM1 for attempting to factorise $(x^2-4)$; check first and last terms |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x^3+3x^2-4x-12}{x^3+5x^2+6x} = \frac{\ldots}{x(x^2+5x+6)}$ | M1 | Takes common factor $x$ from denominator and writes numerator in factors |
| $= \frac{(x+3)(x+2)(x-2)}{x(x+3)(x+2)}$ | dM1 | Further factorises denominator and cancels |
| $= \frac{x-2}{x} = 1 - \frac{2}{x}$ | A1 | No errors or omissions |
---5.
$$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 4 x - 12$$
\begin{enumerate}[label=(\alph*)]
\item Using the factor theorem, explain why $\mathrm { f } ( x )$ is divisible by $( x + 3 )$.
\item Hence fully factorise $\mathrm { f } ( x )$.
\item Show that $\frac { x ^ { 3 } + 3 x ^ { 2 } - 4 x - 12 } { x ^ { 3 } + 5 x ^ { 2 } + 6 x }$ can be written in the form $A + \frac { B } { x }$ where $A$ and $B$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q5 [8]}}