| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Simple Algebraic Fraction Simplification |
| Difficulty | Moderate -0.3 This is a standard C4 question testing routine polynomial techniques: Factor Theorem verification (substitution), polynomial division, algebraic fraction simplification, and partial fractions with an improper fraction. All parts follow textbook procedures with no novel problem-solving required, though part (b) requires recognizing the improper fraction setup. Slightly easier than average due to straightforward calculations. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(f\left(\frac{1}{2}\right) = 2 \times\left(\frac{1}{2}\right)^3 + 3 \times\left(\frac{1}{2}\right) - 18\left(\frac{1}{2}\right) + 8\) | M1 | use of \(\pm\frac{1}{2}\) substituted in \(f(x)\) |
| \(= \frac{1}{4} + \frac{3}{4} - 9 + 8 = 0 \Rightarrow\) factor | A1 | arithmetic seen and conclusion – minimum seen: \(2x\cdot\frac{1}{8} + 3x \cdot\frac{1}{4} - 18x\frac{1}{2} + 8 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (2x-1)(x^2 + 2x - 8)\) | B1 B1 | or \(p = 2, q = -8\) |
| Answer | Marks | Guidance |
|---|---|---|
| numerator correct; attempt to factorise denominator (algebraic fraction not required) | M1 | |
| \(\frac{4x(x+4)}{(2x-1)(x+4)(x-2)} = \frac{4x}{(2x-1)(x-2)}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^2 = A(x+5)(x-3) + B + Cx\) | M1 | any equivalent method using PFs (see alternative method) |
| \(A = 2\) | B1 | |
| \(2A + C = 0\), \(-15A + B = 0\) | M1 | equate coefficients or use 2 values of \(x\) to find \(B\) and \(C\) |
| \(C = -4\), \(B = 30\) | A1 | both \(B\) and \(C\) correct |
| Answer | Marks |
|---|---|
| \(x^2 + 2x - 15 \, \overline{) 2x^2}\) complete division | (M1) |
| \(2x^2 + 4x - 30\) | |
| \(-4x + 30\) | |
| \(A = 2\) | (B1) |
| \(B = 30\) | (A1) |
| \(C = -4\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2x^2}{(x+5)(x-3)} = A + \frac{D}{x+5} + \frac{E}{x-3}\) | ||
| \(2x^2 = A(x+5)(x-3) + D(x-3) + E(x+5)\) | (M1) | find \(D\) and \(E\) |
| \(x = 3\): \(18 = 8E\), \(E = \frac{9}{4}\) | ||
| \(x = -5\): \(50 = -8D\), \(D = -\frac{25}{4}\) | (M1) | |
| \(x = 0, 0 = -15A + \left(-\frac{25}{4}\right)(-3) + \frac{9}{4}(5)\) | (B1) | |
| \(A = 2\) | ||
| \(\frac{D}{x+5} + \frac{E}{x-3} = \frac{-25}{4(x+5)} + \frac{9}{4(x-3)}\) | (M1) | recombine to required form |
| \(= \frac{-25(x-3) + 9(x+5)}{4(x+5)(x-3)} = \frac{120 - 16x}{4(x+5)(x-3)} = \frac{30 - 4x}{(x+5)(x-3)}\) | (A1) | CAO |
**2(a)(i)**
| $f\left(\frac{1}{2}\right) = 2 \times\left(\frac{1}{2}\right)^3 + 3 \times\left(\frac{1}{2}\right) - 18\left(\frac{1}{2}\right) + 8$ | M1 | use of $\pm\frac{1}{2}$ substituted in $f(x)$ |
| $= \frac{1}{4} + \frac{3}{4} - 9 + 8 = 0 \Rightarrow$ factor | A1 | arithmetic seen and conclusion – minimum seen: $2x\cdot\frac{1}{8} + 3x \cdot\frac{1}{4} - 18x\frac{1}{2} + 8 = 0$ |
**2(a)(ii)**
| $f(x) = (2x-1)(x^2 + 2x - 8)$ | B1 B1 | or $p = 2, q = -8$ |
**2(a)(iii)**
| numerator correct; attempt to factorise denominator (algebraic fraction not required) | M1 | |
| $\frac{4x(x+4)}{(2x-1)(x+4)(x-2)} = \frac{4x}{(2x-1)(x-2)}$ | A1 | CAO |
**2(b)**
| $2x^2 = A(x+5)(x-3) + B + Cx$ | M1 | any equivalent method using PFs (see alternative method) |
| $A = 2$ | B1 | |
| $2A + C = 0$, $-15A + B = 0$ | M1 | equate coefficients or use 2 values of $x$ to find $B$ and $C$ |
| $C = -4$, $B = 30$ | A1 | both $B$ and $C$ correct |
**Alternative Method 1:**
| $x^2 + 2x - 15 \, \overline{) 2x^2}$ complete division | (M1) | |
| $2x^2 + 4x - 30$ | | |
| $-4x + 30$ | | |
| $A = 2$ | (B1) | |
| $B = 30$ | (A1) | |
| $C = -4$ | (A1) | |
**Alternative Method 2:**
| $\frac{2x^2}{(x+5)(x-3)} = A + \frac{D}{x+5} + \frac{E}{x-3}$ | | |
| $2x^2 = A(x+5)(x-3) + D(x-3) + E(x+5)$ | (M1) | find $D$ and $E$ |
| $x = 3$: $18 = 8E$, $E = \frac{9}{4}$ | | |
| $x = -5$: $50 = -8D$, $D = -\frac{25}{4}$ | (M1) | |
| $x = 0, 0 = -15A + \left(-\frac{25}{4}\right)(-3) + \frac{9}{4}(5)$ | (B1) | |
| $A = 2$ | | |
| $\frac{D}{x+5} + \frac{E}{x-3} = \frac{-25}{4(x+5)} + \frac{9}{4(x-3)}$ | (M1) | recombine to required form |
| $= \frac{-25(x-3) + 9(x+5)}{4(x+5)(x-3)} = \frac{120 - 16x}{4(x+5)(x-3)} = \frac{30 - 4x}{(x+5)(x-3)}$ | (A1) | CAO |
**Total: 10 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 18 x + 8$.
\begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $( 2 x - 1 )$ is a factor of $\mathrm { f } ( x )$.
\item Write $\mathrm { f } ( x )$ in the form $( 2 x - 1 ) \left( x ^ { 2 } + p x + q \right)$, where $p$ and $q$ are integers.
\item Simplify the algebraic fraction $\frac { 4 x ^ { 2 } + 16 x } { 2 x ^ { 3 } + 3 x ^ { 2 } - 18 x + 8 }$.
\end{enumerate}\item Express the algebraic fraction $\frac { 2 x ^ { 2 } } { ( x + 5 ) ( x - 3 ) }$ in the form $A + \frac { B + C x } { ( x + 5 ) ( x - 3 ) }$, where $A , B$ and $C$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q2 [10]}}