| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two unknowns with show-that step |
| Difficulty | Moderate -0.3 This is a standard C2 Factor/Remainder Theorem question with routine application of techniques. Part (a) uses the remainder theorem with x=1/2, parts (b-c) solve simultaneous equations, and part (d) performs polynomial division. While multi-part with several steps, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | Way 1: Use \(f(1/2)\) or \(f(-1/2)\) and put equal to 30. Stated \(\frac{24}{8} + \frac{1}{4}A - \frac{3}{2} = B = 30\) and \(A + 4B = 114\)* | M1 A1* |
| Way 2: Long division of \(f(x)\) by \((2x-1)\) as far as remainder put = 30. Obtains \(B + \frac{1}{2}A + \frac{3}{2} = 30\) (o.e) and \(A + 4B = 114\)* | M1 A1* | M1: for attempting long division of \(f(x)\) by \((2x-1)\) obtaining \(12x^2 + ...x + ...\) as quotient and remainder term put equal to 30. A1*: Obtaining correct equation correctly |
| (2) | ||
| (b) | Way 1: Used \(f(-1)\) or \(f(1) = 0\). Stated \(-24 + A + 3 + B = 0\) so \(A + B = 21\) | M1 A1 |
| Way 2: Long division of \(f(x)\) by \((x+1)\) as far as remainder put = 0. Obtains \(B - 21 + A = 0\) | M1 A1 | M1: for attempting long division of \(f(x)\) by \((x+1)\) obtaining \(24x^2 + ...x + ...\) as quotient and remainder term put equal to 0 (This may be implied by their equation in part (b)). A1: for obtaining a correct equivalent equation in part (b). (This mark may not be recovered in part (c)). Accept \(A + B = 21\) or \(-A - B = -21\) or \(A + B - 21 = 0\) or \(21 - A - B = 0\) or \(B - 21 + A = 0\) etc. |
| (2) | ||
| (c) | Solves to obtain one of \(A\) or \(B\). Obtains both \(A = -10\) and \(B = 31\) | M1 A1 |
| (2) | ||
| (d) | \(f(x) = (x+1)(24x^2 - 34x + 31)\) or factor is \((24x^2 - 34x + 31)\) | M1A1 |
| (2) | ||
| (8 marks) |
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (a) | **Way 1:** Use $f(1/2)$ or $f(-1/2)$ and put equal to 30. Stated $\frac{24}{8} + \frac{1}{4}A - \frac{3}{2} = B = 30$ and $A + 4B = 114$* | M1 A1* | M1: for attempting either $f(\frac{1}{2})$ or $f(-\frac{1}{2})$ – with numbers substituted into expression and put = 30. A1*: Obtaining correct equation correctly (Signs and powers of ½ need to be simplified correctly) |
| | **Way 2:** Long division of $f(x)$ by $(2x-1)$ as far as remainder put = 30. Obtains $B + \frac{1}{2}A + \frac{3}{2} = 30$ (o.e) and $A + 4B = 114$* | M1 A1* | M1: for attempting long division of $f(x)$ by $(2x-1)$ obtaining $12x^2 + ...x + ...$ as quotient and remainder term put equal to 30. A1*: Obtaining correct equation correctly |
| | | **(2)** | |
| (b) | **Way 1:** Used $f(-1)$ or $f(1) = 0$. Stated $-24 + A + 3 + B = 0$ so $A + B = 21$ | M1 A1 | M1: for calculating $f(-1)$ or $f(1)$ and put equal to 0 (This may be implied by their equation in part (b)). A1: for obtaining a correct equivalent equation in part (b). (This mark may not be recovered in part (c)) |
| | **Way 2:** Long division of $f(x)$ by $(x+1)$ as far as remainder put = 0. Obtains $B - 21 + A = 0$ | M1 A1 | M1: for attempting long division of $f(x)$ by $(x+1)$ obtaining $24x^2 + ...x + ...$ as quotient and remainder term put equal to 0 (This may be implied by their equation in part (b)). A1: for obtaining a correct equivalent equation in part (b). (This mark may not be recovered in part (c)). Accept $A + B = 21$ or $-A - B = -21$ or $A + B - 21 = 0$ or $21 - A - B = 0$ or $B - 21 + A = 0$ etc. |
| | | **(2)** | |
| (c) | Solves to obtain one of $A$ or $B$. Obtains both $A = -10$ and $B = 31$ | M1 A1 | M1: Eliminate one variable and solve to obtain $A$ or $B$. A1: Both correct |
| | | **(2)** | |
| (d) | $f(x) = (x+1)(24x^2 - 34x + 31)$ or factor is $(24x^2 - 34x + 31)$ | M1A1 | M1: Uses their values of $A$ and $B$ in given cubic (even wrong way round) and attempts to divide by $(x + 1)$ leading to 3TO beginning with $24x^2$ and including an $x$ term and constant term. This may be done by variety of methods including long division, comparison of coefficients, inspection etc. (If values of $A$ and $B$ were wrong there may be remainder but this may be ignored). If they used division in part (b) they may substitute $A$ and $B$ into their quotient expression from (b). A1: $24x^2 - 34x + 31$.... Credit when seen and use isw if miscopied later or if attempt made to solve. |
| | | **(2)** | |
| | | **(8 marks)** | |
---3.
$$f ( x ) = 24 x ^ { 3 } + A x ^ { 2 } - 3 x + B$$
where $A$ and $B$ are constants.\\
When $\mathrm { f } ( x )$ is divided by $( 2 x - 1 )$ the remainder is 30
\begin{enumerate}[label=(\alph*)]
\item Show that $A + 4 B = 114$
Given also that ( $x + 1$ ) is a factor of $\mathrm { f } ( x )$,
\item find another equation in $A$ and $B$.
\item Find the value of $A$ and the value of $B$.
\item Hence find a quadratic factor of $\mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2018 Q3 [8]}}