| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| 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 factor theorem question with routine steps: substitute x=2 to verify the factor (part a), then perform polynomial division or comparison to find the quadratic factor, followed by standard factorisation. It requires only direct application of well-practiced techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts \(f(2)\) or \(f(-2)\) | M1 | |
| \(f(2) = 2(2)^3 - 7(2)^2 + 4(2) + 4 = 0\), and so \((x-2)\) is a factor | A1 | \(f(2) = 0\) with no sign or substitution errors; must state conclusion e.g. "hence factor" or "it is a factor". Not \(= 0\) just underlined and not hence \((2\) or \(f(2))\) is a factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = \{(x-2)\}(2x^2 - 3x - 2)\) | M1 A1 | M1: Attempts long division by \((x-2)\) or other method to obtain \((2x^2 \pm ax \pm b)\), \(a \neq 0\), even with remainder. A1: \((2x^2 - 3x - 2)\) |
| \(= (x-2)(x-2)(2x+1)\) or \((x-2)^2(2x+1)\) or equivalent | dM1 A1 | dM1: Factorises 3-term quadratic — dependent on previous M1, no remainder. A1: cao — needs all three factors on one line. Ignore solutions to quadratic equation |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts $f(2)$ or $f(-2)$ | M1 | |
| $f(2) = 2(2)^3 - 7(2)^2 + 4(2) + 4 = 0$, and so $(x-2)$ is a factor | A1 | $f(2) = 0$ with no sign or substitution errors; must state conclusion e.g. "hence factor" or "it is a factor". **Not** $= 0$ just underlined and **not** hence $(2$ or $f(2))$ is a factor |
**Note:** Long division scores no marks in part (a). The factor theorem is required.
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = \{(x-2)\}(2x^2 - 3x - 2)$ | M1 A1 | M1: Attempts long division by $(x-2)$ or other method to obtain $(2x^2 \pm ax \pm b)$, $a \neq 0$, even with remainder. A1: $(2x^2 - 3x - 2)$ |
| $= (x-2)(x-2)(2x+1)$ or $(x-2)^2(2x+1)$ or equivalent | dM1 A1 | dM1: Factorises 3-term quadratic — dependent on previous M1, no remainder. A1: cao — needs all three factors **on one line**. Ignore solutions to quadratic equation |
**Note:** $= (x-2)(\frac{1}{2}x - 1)(4x+2)$ loses last mark as not **fully** factorised. For correct answers only award full marks in (b).
---2.
$$f ( x ) = 2 x ^ { 3 } - 7 x ^ { 2 } + 4 x + 4$$
\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.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q2 [6]}}