| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 9 |
| 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 Factor and Remainder Theorem question with routine algebraic manipulation. Part (a) is direct application of the remainder theorem, part (b) involves solving simultaneous equations, and part (c) is straightforward factorization once constants are known. Slightly easier than average due to its predictable structure and minimal problem-solving required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2(1)^3 - 3(1)^2 + p(1) + q = -6\) | M1 | Attempts \(f(\pm 1) = -6\) |
| \(p + q = -5\) | A1 | Correct equation with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Long division by \((x-1)\) leading to remainder in \(p\) and \(q\) set \(= -6\) | M1 | Attempts long division correctly (allow sign slips only) leading to remainder in \(p\) and \(q\) which is set \(= -6\) |
| \(p + q = -5\) | A1 | Correct equation with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2(-2)^3 - 3(-2)^2 + p(-2) + q = 0\) | M1 | Clear attempt at \(f(-2) = 0\) or \(f(2) = 0\). May be implied by correct equation, but if equation is incorrect and no method shown score M0 |
| \(p + q = -5,\ q - 2p = 28 \Rightarrow p = -11,\ q = 6\) | M1 | Solves simultaneously. Must be using \(p+q=-5\) and their linear equation in \(p\) and \(q\), must reach values for both \(p\) and \(q\) |
| \(p = -11,\ q = 6\) | A1 | Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\frac{2x^3-3x^2-11x+6}{x+2} = 2x^2+kx+\ldots\) | M1 | Divides \(f(x)\) by \((x+2)\) or compares coefficients or uses inspection and obtains at least the first 2 terms of a quadratic with \(2x^2\) as the first term and an \(x\) term. Must be seen in (c). |
| \(2x^2-7x+3\) | A1 | Correct quadratic |
| \(2x^2-7x+3=(2x-1)(x-3)\) | dM1 | Attempts to factorise their 3 term quadratic expression. Usual rules apply here so if \(2x^2-7x+3\) is factorised as \((x-\frac{1}{2})(x-3)\), this scores M0 unless the factor of 2 appears later. Dependent on the first M mark. |
| \(f(x)=(x+2)(2x-1)(x-3)\) or \(f(x)=2(x+2)(x-\frac{1}{2})(x-3)\) | A1 | Fully correct factorisation. Must see all factors together on one line and no commas in between. |
# Question 8:
## Parts (a) and (b) marked together:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(1)^3 - 3(1)^2 + p(1) + q = -6$ | M1 | Attempts $f(\pm 1) = -6$ |
| $p + q = -5$ | A1 | Correct equation with no errors |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Long division by $(x-1)$ leading to remainder in $p$ and $q$ set $= -6$ | M1 | Attempts long division correctly (allow sign slips only) leading to remainder in $p$ and $q$ which is set $= -6$ |
| $p + q = -5$ | A1 | Correct equation with no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(-2)^3 - 3(-2)^2 + p(-2) + q = 0$ | M1 | Clear attempt at $f(-2) = 0$ or $f(2) = 0$. May be implied by correct equation, but if equation is incorrect and no method shown score M0 |
| $p + q = -5,\ q - 2p = 28 \Rightarrow p = -11,\ q = 6$ | M1 | Solves simultaneously. Must be using $p+q=-5$ and their linear equation in $p$ and $q$, must reach values for both $p$ and $q$ |
| $p = -11,\ q = 6$ | A1 | Correct values |
## Question 8(c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\frac{2x^3-3x^2-11x+6}{x+2} = 2x^2+kx+\ldots$ | M1 | Divides $f(x)$ by $(x+2)$ or compares coefficients or uses inspection and obtains at least the first 2 terms of a quadratic with $2x^2$ as the first term and an $x$ term. Must be seen in (c). |
| $2x^2-7x+3$ | A1 | Correct quadratic |
| $2x^2-7x+3=(2x-1)(x-3)$ | dM1 | Attempts to factorise their 3 term quadratic expression. Usual rules apply here so if $2x^2-7x+3$ is factorised as $(x-\frac{1}{2})(x-3)$, this scores M0 unless the factor of 2 appears later. Dependent on the first M mark. |
| $f(x)=(x+2)(2x-1)(x-3)$ or $f(x)=2(x+2)(x-\frac{1}{2})(x-3)$ | A1 | Fully correct factorisation. Must see all factors together on one line and no commas in between. |
---8.
$$f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } + p x + q$$
where $p$ and $q$ are constants.\\
When $\mathrm { f } ( x )$ is divided by $( x - 1 )$, the remainder is - 6
\begin{enumerate}[label=(\alph*)]
\item Use the remainder theorem to show that $p + q = - 5$
Given also that $( x + 2 )$ is a factor of $\mathrm { f } ( x )$,
\item find the value of $p$ and the value of $q$.
\item Factorise $\mathrm { f } ( \mathrm { x } )$ completely.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q8 [9]}}