| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| 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=1), factor theorem (show f(-1)=0), and polynomial factorisation using standard techniques. All steps are routine C2-level procedures with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(1)\) or \(f(-1)\) | M1 | Can be implied; only one slip permitted. Long division attempt also acceptable |
| \(f(1) = 2 - 7 - 5 + 4 = -6\) | A1 | Award M1A1 for \(-6\) without working. Award A0 if candidate finds \(-6\) but states remainder is 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(-1)\) | M1 | Long division scores no marks in part (b); factor theorem required |
| \(f(-1) = 2(-1)^3 - 7(-1)^2 - 5(-1) + 4 = 0\), so \((x+1)\) is a factor | A1 | Must correctly show \(f(-1)=0\) with no sign/substitution errors and give conclusion. "Hence factor", "tick", or "QED" acceptable as conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = (x+1)(2x^2 - 9x + 4)\) | M1 A1 | 1st M1: attempts long division or other method to obtain \((2x^2 \pm ax \pm b)\), \(a \neq 0\). Working need not be shown ("by inspection" allowed). Must appear in part (c) only |
| \(= (x+1)(2x-1)(x-4)\) | dM1 A1 | 2nd dM1: factorises 3-term quadratic (dependent on previous M). 2nd A1: cao, all three factors on one line |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(1)$ or $f(-1)$ | M1 | Can be implied; only one slip permitted. Long division attempt also acceptable |
| $f(1) = 2 - 7 - 5 + 4 = -6$ | A1 | Award M1A1 for $-6$ without working. Award A0 if candidate finds $-6$ but states remainder is 6 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(-1)$ | M1 | Long division scores no marks in part (b); factor theorem required |
| $f(-1) = 2(-1)^3 - 7(-1)^2 - 5(-1) + 4 = 0$, so $(x+1)$ is a factor | A1 | Must correctly show $f(-1)=0$ with no sign/substitution errors **and** give conclusion. "Hence factor", "tick", or "QED" acceptable as conclusion |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = (x+1)(2x^2 - 9x + 4)$ | M1 A1 | 1st M1: attempts long division or other method to obtain $(2x^2 \pm ax \pm b)$, $a \neq 0$. Working need not be shown ("by inspection" allowed). Must appear in part (c) only |
| $= (x+1)(2x-1)(x-4)$ | dM1 A1 | 2nd dM1: factorises 3-term quadratic (dependent on previous M). 2nd A1: cao, all three factors on one line |
**Alternative (first two marks):** 1st M1: Expands $(x+1)(2x^2+ax+b)$ then compares coefficients to find $a$ and $b$. 1st A1: $a=-9$, $b=4$
**Note:** Some candidates going from $\{(x+1)\}(2x^2-9x+4)$ to roots only: award M1A1M1A0
---1.
$$f ( x ) = 2 x ^ { 3 } - 7 x ^ { 2 } - 5 x + 4$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by $( x - 1 )$.
\item Use the factor theorem to show that ( $x + 1$ ) is a factor of $\mathrm { f } ( x )$.
\item Factorise f(x) completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2011 Q1 [8]}}