| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find remainder(s) then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the Remainder Theorem requiring direct substitution to find remainders, followed by factorization once a root is identified. The question is routine for C2 level with clear signposting ('hence') and involves standard techniques with no problem-solving insight required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(f(3) = 3^3 - 2 \times 3^2 - 4 \times 3 + 8 = 5\) | M1; A1 | |
| (a)(ii) \(f(-2) = (-8 - 8 + 8 + 8) = 0\) | A1 (3) | M1 is for attempt at either \(f(3)\) or \(f(-3)\) or \(f(-2)\) or \(f(2)\) in part (ii). (B1 on Epen, but A1 in fact) |
| (b) \([(x+2)](x^2 - 4x + 4) = (x+2)(x-2)^2\) | M1 A1; M1 | (= 0 not required) [must be seen or used in (b)] |
| Solutions: \(x = 2\) or \(x = -2\) (both) or \((-2, 2, 2)\) | A1 (4) |
| Answer | Marks |
|---|---|
| - Alternative (Long division): Divide by \((x-3)\) OR \((x+2)\) to get \(x^2 + ax + b\) where \(a\) may be zero | [M1] |
| - \(x^2 + x - 1\) and \(+5\) seen i.s.w. (or "remainder = 5") | [A1] |
| - \(x^2 - 4x + 4\) and \(0\) seen (or "no remainder") | [B1] |
| Answer | Marks | Guidance |
|---|---|---|
| - [Usual rule: \(x^2 + ax + b = (x+c)(x+d)\) where \( | cd | = |
| Answer | Marks | Guidance |
|---|---|---|
| - Alternative (first two marks): \((x+2)(x^2 + bx + c) = x^3 + (2+b)x^2 + (2b+c)x + 2c = 0\) and then compare with \(x^3 - 2x^2 - 4x + 8 = 0\) to find \(b\) and \(c\) | [M1] | |
| - \(b = -4, c = 4\) | [A1] | |
| - Method of grouping: \(x^3 - 2x^2 - 4x + 8 = x^2(x-2), 4(x+2)\) | M1; \(= x^2(x-2) - 4(x-2)\) | A1 |
| - \([(x^2-4)(x-2)] = (x+2)(x-2)^2\) | M1 | |
| - Solutions: \(x = 2, x = -2\) both | A1 |
**(a)(i)** $f(3) = 3^3 - 2 \times 3^2 - 4 \times 3 + 8 = 5$ | M1; A1 |
**(a)(ii)** $f(-2) = (-8 - 8 + 8 + 8) = 0$ | A1 (3) | M1 is for attempt at either $f(3)$ or $f(-3)$ or $f(-2)$ or $f(2)$ in part (ii). (B1 on Epen, but A1 in fact)
**(b)** $[(x+2)](x^2 - 4x + 4) = (x+2)(x-2)^2$ | M1 A1; M1 | (= 0 not required) [must be seen or used in (b)] | M1 |
Solutions: $x = 2$ or $x = -2$ (both) or $(-2, 2, 2)$ | A1 (4) |
**Notes:**
- Alternative (Long division): Divide by $(x-3)$ OR $(x+2)$ to get $x^2 + ax + b$ where $a$ may be zero | [M1]
- $x^2 + x - 1$ and $+5$ seen i.s.w. (or "remainder = 5") | [A1]
- $x^2 - 4x + 4$ and $0$ seen (or "no remainder") | [B1]
- First M1 requires division by a found factor; e.g. $(x+2)$, $(x-2)$ or what candidate thinks is a factor to get $(x^2 + ax + b)$ where $a$ may be zero
- First A1 for $[(x+2)](x^2 - 4x + 4)$ or $(x-2)[(x^2 - 4)]$
- Second M1 attempt to factorise their found quadratic (or use formula correctly)
- [Usual rule: $x^2 + ax + b = (x+c)(x+d)$ where $|cd| = |b|$]
- N.B. Second A1 is for solutions, not factors
- Alternative (first two marks): $(x+2)(x^2 + bx + c) = x^3 + (2+b)x^2 + (2b+c)x + 2c = 0$ and then compare with $x^3 - 2x^2 - 4x + 8 = 0$ to find $b$ and $c$ | [M1]
- $b = -4, c = 4$ | [A1]
- Method of grouping: $x^3 - 2x^2 - 4x + 8 = x^2(x-2), 4(x+2)$ | M1; $= x^2(x-2) - 4(x-2)$ | A1
- $[(x^2-4)(x-2)] = (x+2)(x-2)^2$ | M1
- Solutions: $x = 2, x = -2$ both | A1
---\begin{enumerate}[label=(\alph*)]
\item Find the remainder when
$$x^3 - 2x^2 - 4x + 8$$
is divided by
\begin{enumerate}[label=(\roman*)]
\item $x - 3$,
\item $x + 2$.
\end{enumerate}
[3]
\item Hence, or otherwise, find all the solutions to the equation
$$x^3 - 2x^2 - 4x + 8 = 0.$$
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q1 [7]}}