| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing routine application of the factor theorem and polynomial division. Part (a) requires simple substitution of x=-2, part (b) is standard algebraic long division followed by factorising a quadratic, and part (c) is direct reading of roots from the factorised form. No problem-solving insight needed, just mechanical application of well-practiced techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6\) | M1 | Attempting \(f(\pm 2)\); no \(x\)s; allow invisible brackets |
| \(= -8 + 16 - 2 - 6 = 0\), \(\therefore x+2\) is a factor | A1 (2) | Must state \(= 0\) and minimal conclusion; long division scores M0 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^3 + 4x^2 + x - 6 = (x+2)(x^2 + 2x - 3)\) | M1, A1 | Division by \((x+2)\) to get \(x^2+ax+b\) where \(a\neq 0\), \(b\neq 0\) |
| \(= (x+2)(x+3)(x-1)\) | M1, A1 (4) | Attempt to factorise quadratic; combining all 3 factors not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-3, -2, 1\) | B1 (1) | Answer to (b) can be seen in (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Show \(f(-3) = 0\); allow invisible brackets | M1 | |
| \(\therefore x + 3\) is a factor | A1 | |
| Show \(f(1) = 0\) | M1 | |
| \(\therefore x - 1\) is a factor | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-3, -2, 1\) or \((-3, 0), (-2, 0), (1, 0)\) only | B1 | Do not ignore subsequent working. Ignore any working in previous parts. Can be seen in (b) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6$ | M1 | Attempting $f(\pm 2)$; no $x$s; allow invisible brackets |
| $= -8 + 16 - 2 - 6 = 0$, $\therefore x+2$ is a factor | A1 **(2)** | Must state $= 0$ and minimal conclusion; long division scores M0 A0 |
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^3 + 4x^2 + x - 6 = (x+2)(x^2 + 2x - 3)$ | M1, A1 | Division by $(x+2)$ to get $x^2+ax+b$ where $a\neq 0$, $b\neq 0$ |
| $= (x+2)(x+3)(x-1)$ | M1, A1 **(4)** | Attempt to factorise quadratic; combining all 3 factors not required |
## Question 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-3, -2, 1$ | B1 **(1)** | Answer to (b) can be seen in (c) |
## Question 5(b) Alternative (Factor Theorem):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Show $f(-3) = 0$; allow invisible brackets | M1 | |
| $\therefore x + 3$ is a factor | A1 | |
| Show $f(1) = 0$ | M1 | |
| $\therefore x - 1$ is a factor | A1 | |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-3, -2, 1$ or $(-3, 0), (-2, 0), (1, 0)$ only | B1 | Do not ignore subsequent working. Ignore any working in previous parts. Can be seen in (b) |
---5.
$$f ( x ) = x ^ { 3 } + 4 x ^ { 2 } + x - 6$$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( x + 2 )$ is a factor of $\mathrm { f } ( x )$.
\item Factorise $\mathrm { f } ( x )$ completely.
\item Write down all the solutions to the equation
$$x ^ { 3 } + 4 x ^ { 2 } + x - 6 = 0$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q5 [7]}}