| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| 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 (substitute x=-2), factor theorem (show f(-3)=0), followed by polynomial division and factorising a quadratic. All steps are routine C2 techniques with no problem-solving insight required, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(-2) = 2(-2)^3 + 3(-2)^2 - 29(-2) - 60 = -16 + 12 + 58 - 60 = -6\) | M1, A1 | M: Attempt \(f(2)\) or \(f(-2)\). (2 marks) |
| (b) \(f(-3) = 2(-3)^3 + 3(-3)^2 - 29(-3) - 60 = -54 + 27 + 87 - 60 = 0\) | M1, A1 | M: Attempt \(f(3)\) or \(f(-3)\). Therefore \((x + 3)\) is a factor. (2 marks) |
| (c) \((x + 3)(2x^2 - 3x - 20) = (x + 3)(2x + 5)(x - 4)\) | M1, A1, M1, A1 | First M requires division by \((x + 2)\) to get \((2x^2 + ax + b)\), \(a \neq 0, b \neq 0\). Second M for attempt to factorise their quadratic. Usual rule: \((2x^2 + ax + b) = (2x + c)(x + d)\), where \(\ |
| (8 marks) | Alternative (first 2 marks): \((x + 3)(2x^2 + ax + b) = 2x^3 + (6 + a)x^2 + (3a + b)x + 3b = 0\), then compare coefficients to find values of \(a\) and \(b\). \(a = -3, b = -20\). Alternative: Factor theorem: Finding that \(f\left(-\frac{5}{2}\right) = 0 \therefore\) factor is \((2x + 5)\). Finding that \(f(4) = 0 \therefore\) factor is \((x - 4)\). "Combining" all 3 factors is not required. If just one of these is found, score the first 2 marks M1 A1 M0 A0. Losing a factor of 2: \((x + 3)\left(x + \frac{5}{2}\right)(x - 4)\) scores M1 A1 M1 A0. Answer only, one sign wrong; e.g. \((x + 3)(2x - 5)(x - 4)\) scores M1 A1 M1 A0. |
**(a)** $f(-2) = 2(-2)^3 + 3(-2)^2 - 29(-2) - 60 = -16 + 12 + 58 - 60 = -6$ | M1, A1 | M: Attempt $f(2)$ or $f(-2)$. (2 marks) |
| --- | --- | --- |
| **(b)** $f(-3) = 2(-3)^3 + 3(-3)^2 - 29(-3) - 60 = -54 + 27 + 87 - 60 = 0$ | M1, A1 | M: Attempt $f(3)$ or $f(-3)$. Therefore $(x + 3)$ is a factor. (2 marks) |
| **(c)** $(x + 3)(2x^2 - 3x - 20) = (x + 3)(2x + 5)(x - 4)$ | M1, A1, M1, A1 | First M requires division by $(x + 2)$ to get $(2x^2 + ax + b)$, $a \neq 0, b \neq 0$. Second M for attempt to factorise their quadratic. Usual rule: $(2x^2 + ax + b) = (2x + c)(x + d)$, where $\|cd\| = \|b\|$. (4 marks) |
| | (8 marks) | **Alternative (first 2 marks):** $(x + 3)(2x^2 + ax + b) = 2x^3 + (6 + a)x^2 + (3a + b)x + 3b = 0$, then compare coefficients to find values of $a$ and $b$. $a = -3, b = -20$. **Alternative:** Factor theorem: Finding that $f\left(-\frac{5}{2}\right) = 0 \therefore$ factor is $(2x + 5)$. Finding that $f(4) = 0 \therefore$ factor is $(x - 4)$. "Combining" all 3 factors is not required. If just one of these is found, score the first 2 marks M1 A1 M0 A0. Losing a factor of 2: $(x + 3)\left(x + \frac{5}{2}\right)(x - 4)$ scores M1 A1 M1 A0. Answer only, one sign wrong; e.g. $(x + 3)(2x - 5)(x - 4)$ scores M1 A1 M1 A0. |$$f ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 29 x - 60$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by $( x + 2 )$.
\item Use the factor theorem to show that $( x + 3 )$ is a factor of $\mathrm { f } ( x )$.
\item Factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q4 [8]}}