| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Partial Fraction Form via Division |
| Difficulty | Standard +0.3 This is a standard C3 question combining polynomial division, partial fractions, and numerical methods. Part (a) is routine algebraic division with straightforward partial fraction decomposition. Part (b) requires basic curve sketching. Part (c) applies a given iterative formula—no derivation needed. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02q Use intersection points: of graphs to solve equations1.02y Partial fractions: decompose rational functions1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Polynomial division: \(x^2 + 6\) into \(x^4 + x^3 - 5x^2 + 0x - 9\) | M1 A1 | |
| Quotient: \(x^2 + 0x + 1\), remainder: \(-x - 3\) | M1 A1 | |
| \(\therefore f(x) = x^2 + 1 + \frac{-x - 3}{x^2 + x - 6}\) | A1 | |
| \(= x^2 + 1 - \frac{x + 3}{(x - 2)(x + 3)} = x^2 + 1 - \frac{1}{x - 2}\) | M1 A1 | \([A = 1, B = -1, C = -2]\) |
| (b) Sketch showing \(y = x^2 + 1\) and \(y = \frac{1}{x - 2}\) | B2 | |
| (c) \(f(x) = 0 \Rightarrow x^2 + 1 = \frac{1}{x - 2}\), graphs intersect once \(\therefore\) exactly one real root | B1 | |
| E.g. \(x_0 = 3\), \(x_1 = 2.1\), \(x_2 = 2.1848\), \(x_3 = 2.1732\), \(x_4 = 2.1747\) | M1 A1 | |
| \(\therefore\) root \(= 2.17\) (3sf) | A1 | |
| \(f(2.165) = -0.37\), \(f(2.175) = 0.016\) | M1 | |
| Sign change, \(f(x)\) continuous \(\therefore\) root | A1 | (13 marks) |
**(a)** Polynomial division: $x^2 + 6$ into $x^4 + x^3 - 5x^2 + 0x - 9$ | M1 A1 |
Quotient: $x^2 + 0x + 1$, remainder: $-x - 3$ | M1 A1 |
$\therefore f(x) = x^2 + 1 + \frac{-x - 3}{x^2 + x - 6}$ | A1 |
$= x^2 + 1 - \frac{x + 3}{(x - 2)(x + 3)} = x^2 + 1 - \frac{1}{x - 2}$ | M1 A1 | $[A = 1, B = -1, C = -2]$
**(b)** Sketch showing $y = x^2 + 1$ and $y = \frac{1}{x - 2}$ | B2 |
**(c)** $f(x) = 0 \Rightarrow x^2 + 1 = \frac{1}{x - 2}$, graphs intersect once $\therefore$ exactly one real root | B1 |
E.g. $x_0 = 3$, $x_1 = 2.1$, $x_2 = 2.1848$, $x_3 = 2.1732$, $x_4 = 2.1747$ | M1 A1 |
$\therefore$ root $= 2.17$ (3sf) | A1 |
$f(2.165) = -0.37$, $f(2.175) = 0.016$ | M1 |
Sign change, $f(x)$ continuous $\therefore$ root | A1 | (13 marks)\begin{enumerate}
\item $f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 5 x ^ { 2 } - 9 } { x ^ { 2 } + x - 6 }$.\\
(a) Using algebraic division, show that
\end{enumerate}
$$f ( x ) = x ^ { 2 } + A + \frac { B } { x + C }$$
where $A , B$ and $C$ are integers to be found.\\
(b) By sketching two suitable graphs on the same set of axes, show that the equation $\mathrm { f } ( x ) = 0$ has exactly one real root.\\
(c) Use the iterative formula
$$x _ { n + 1 } = 2 + \frac { 1 } { x _ { n } ^ { 2 } + 1 } ,$$
with a suitable starting value to find the root of the equation $\mathrm { f } ( x ) = 0$ correct to 3 significant figures and justify the accuracy of your answer.
\hfill \mbox{\textit{Edexcel C3 Q8 [13]}}