| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.3 This is a straightforward application of the Remainder and Factor Theorems requiring substitution of x=2 and x=-3 to create two simultaneous equations in a and b. The algebraic manipulation is routine for C2 level, making it slightly easier than average but not trivial since it requires coordinating two conditions and solving simultaneous equations. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(2) = 2(2)^3 + (2)^2 + 2a + b = 25\) | M1 | \(f(\pm 2) = 25\) |
| \(16 + 4 + 2a + b = 25 \Rightarrow 2a + b = 5\) * | A1 | Correct completion to printed answer. If \(f(2)\) not seen explicitly and "\(16+4+2a+b=25\)" is incorrect, score M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x^3 + x^2 + ax + b \div (x-2)\); Quotient \(= 2x^2 + 5x + a + 10\); Remainder \(= 2a + b + 20\) | M1 | Attempt Quotient & Remainder. Needs quotient of form \(2x^2 + kx + f(a)\) and remainder that is a function of \(a\) and \(b\) |
| \(2a + b + 20 = 25 \Rightarrow 2a + b = 5\) * | A1 | Correct completion to printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-3) = 2(-3)^3 + (-3)^2 - 3a + b = 0\) | M1 | \(f(\pm 3) = 0\) |
| \(2a + b = 5\), \(b - 3a = 45 \rightarrow a =\) or \(b =\) | M1 | Solves simultaneously to \(a=\) or \(b=\) |
| \(a = -8\), \(b = 21\) | A1, A1 | First A1: One correct constant. Second A1: Both constants correct |
## Question 2:
### Part (a) Way 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(2) = 2(2)^3 + (2)^2 + 2a + b = 25$ | M1 | $f(\pm 2) = 25$ |
| $16 + 4 + 2a + b = 25 \Rightarrow 2a + b = 5$ * | A1 | Correct completion to printed answer. If $f(2)$ not seen explicitly and "$16+4+2a+b=25$" is incorrect, score M0 |
### Part (a) Way 2 (Long Division):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x^3 + x^2 + ax + b \div (x-2)$; Quotient $= 2x^2 + 5x + a + 10$; Remainder $= 2a + b + 20$ | M1 | Attempt Quotient & Remainder. Needs quotient of form $2x^2 + kx + f(a)$ and remainder that is a function of $a$ and $b$ |
| $2a + b + 20 = 25 \Rightarrow 2a + b = 5$ * | A1 | Correct completion to printed answer |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-3) = 2(-3)^3 + (-3)^2 - 3a + b = 0$ | M1 | $f(\pm 3) = 0$ |
| $2a + b = 5$, $b - 3a = 45 \rightarrow a =$ or $b =$ | M1 | Solves simultaneously to $a=$ or $b=$ |
| $a = -8$, $b = 21$ | A1, A1 | First A1: One correct constant. Second A1: Both constants correct |
---2. $\mathrm { f } ( x ) = 2 x ^ { 3 } + x ^ { 2 } + a x + b$, where $a$ and $b$ are constants.
Given that when $\mathrm { f } ( x )$ is divided by ( $x - 2$ ) the remainder is 25 ,
\begin{enumerate}[label=(\alph*)]
\item show that $2 a + b = 5$
Given also that $( x + 3 )$ is a factor of $\mathrm { f } ( x )$,
\item find the value of $a$ and the value of $b$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e7043e7a-2c8f-425a-8471-f647828cc297-05_90_97_2613_1784}\\
\includegraphics[max width=\textwidth, alt={}, center]{e7043e7a-2c8f-425a-8471-f647828cc297-05_52_169_2709_1765}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q2 [6]}}