| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with equal remainders |
| Difficulty | Moderate -0.3 This is a straightforward application of the Remainder Theorem requiring students to set up two equations from equal remainders (f(2) = f(-1)) and one equation from the factor condition (f(-3) = 0), then solve for constants a and b. The algebraic manipulation is routine and the question follows a standard textbook pattern, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(2) = 16 + 40 + 2a + b\) or \(f(-1) = 1 - 5 - a + b\) | M1 A1 | Attempts \(f(\pm 2)\) or \(f(\pm 1)\) |
| Finds 2nd remainder and equates: \(16 + 40 + 2a + b = 1 - 5 - a + b\) | M1 A1 | Attempts other remainder and equates |
| \(a = -20\) | A1 cso | Correct equation in \(a\) and \(b\), then A1 for \(a = -20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-3) = (-3)^4 + 5(-3)^3 - 3a + b = 0\) | M1 A1ft | Puts \(f(\pm 3) = 0\); A1 for \(f(-3)=0\) with no sign/substitution errors |
| \(81 - 135 + 60 + b = 0\) gives \(b = -6\) | A1 cso | Follow through on \(a\); \(b = -6\) is cso |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(2) = 16 + 40 + 2a + b$ **or** $f(-1) = 1 - 5 - a + b$ | M1 A1 | Attempts $f(\pm 2)$ or $f(\pm 1)$ |
| Finds 2nd remainder and equates: $16 + 40 + 2a + b = 1 - 5 - a + b$ | M1 A1 | Attempts other remainder and equates |
| $a = -20$ | A1 cso | Correct equation in $a$ and $b$, then A1 for $a = -20$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-3) = (-3)^4 + 5(-3)^3 - 3a + b = 0$ | M1 A1ft | Puts $f(\pm 3) = 0$; A1 for $f(-3)=0$ with no sign/substitution errors |
| $81 - 135 + 60 + b = 0$ gives $b = -6$ | A1 cso | Follow through on $a$; $b = -6$ is cso |
---6.
$$f ( x ) = x ^ { 4 } + 5 x ^ { 3 } + a x + b$$
where $a$ and $b$ are constants.
The remainder when $\mathrm { f } ( x )$ is divided by ( $x - 2$ ) is equal to the remainder when $\mathrm { f } ( x )$ is divided by $( x + 1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
Given that $( x + 3 )$ is a factor of $\mathrm { f } ( x )$,
\item find the value of $b$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q6 [8]}}