| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor and Remainder Theorems requiring substitution of x=1 and x=-2, then solving two simultaneous linear equations. It's a standard C2 textbook exercise with clear signposting and routine algebraic manipulation, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1) = a + b - 4 - 3 = 0\) or \(a + b - 7 = 0\) | M1 | Attempt \(f(\pm 1)\) |
| \(a + b = 7\) | A1 | Must be \(f(1)\) and \(= 0\) needs to be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(-2) = a(-2)^3 + b(-2)^2 - 4(-2) - 3 = 9\) | M1 | Attempt \(f(\pm 2)\) and uses \(f(\pm 2) = 9\) |
| \(-8a + 4b + 8 - 3 = 9\) | A1 | Correct equation with exponents of \((-2)\) removed |
| Solves the given equation from part (a) and their equation in \(a\) and \(b\) from part (b) as far as \(a =\) ... or \(b =\) ... | M1 | |
| \(a = 2\) and \(b = 5\) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((ax^3 + bx^2 - 4x - 3) \div (x-1) = ax^2 + px + q\), where \(p\) and \(q\) are in terms of \(a\) or \(b\) or both, and sets remainder \(= 0\). NB Quotient \(= ax^2 + (a+b)x + (a+b-4)\) | M1 | |
| \(a + b = 7\) | A1 | |
| \((ax^3 + bx^2 - 4x - 3) \div (x+2) = ax^2 + px + q\), where \(p\) and \(q\) are in terms of \(a\) or \(b\) or both, and sets remainder \(= 9\). NB Quotient \(= ax^2 + (b-2a)x + (4a - 4 - 2b)\) | M1 | |
| \(4b - 8a + 5 = 9\) | A1 | Follow scheme for final 2 marks |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = a + b - 4 - 3 = 0$ or $a + b - 7 = 0$ | M1 | Attempt $f(\pm 1)$ |
| $a + b = 7$ | A1 | Must be $f(1)$ and $= 0$ needs to be seen |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(-2) = a(-2)^3 + b(-2)^2 - 4(-2) - 3 = 9$ | M1 | Attempt $f(\pm 2)$ and uses $f(\pm 2) = 9$ |
| $-8a + 4b + 8 - 3 = 9$ | A1 | Correct equation with exponents of $(-2)$ removed |
| Solves the given equation from part (a) and their equation in $a$ and $b$ from part (b) as far as $a =$ ... or $b =$ ... | M1 | |
| $a = 2$ and $b = 5$ | A1 | Both correct |
**Long Division Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(ax^3 + bx^2 - 4x - 3) \div (x-1) = ax^2 + px + q$, where $p$ and $q$ are in terms of $a$ or $b$ or both, and sets remainder $= 0$. NB Quotient $= ax^2 + (a+b)x + (a+b-4)$ | M1 | |
| $a + b = 7$ | A1 | |
| $(ax^3 + bx^2 - 4x - 3) \div (x+2) = ax^2 + px + q$, where $p$ and $q$ are in terms of $a$ or $b$ or both, and sets remainder $= 9$. NB Quotient $= ax^2 + (b-2a)x + (4a - 4 - 2b)$ | M1 | |
| $4b - 8a + 5 = 9$ | A1 | Follow scheme for final 2 marks |
---2. $\mathrm { f } ( x ) = a x ^ { 3 } + b x ^ { 2 } - 4 x - 3$, where $a$ and $b$ are constants.
Given that $( x - 1 )$ is a factor of $\mathrm { f } ( x )$,
\begin{enumerate}[label=(\alph*)]
\item show that
$$a + b = 7$$
Given also that, when $\mathrm { f } ( x )$ is divided by $( x + 2 )$, the remainder is 9 ,
\item find the value of $a$ and the value of $b$, showing each step in your working.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q2 [6]}}