| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single unknown from factor condition |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard Factor Theorem application, polynomial division, discriminant analysis, and basic transformations. Part (a) is routine substitution, (b) is straightforward polynomial division, (c) uses the discriminant of a quadratic (standard technique), and (d) applies basic transformations. While it has multiple parts (5 marks total based on typical schemes), each individual step is textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(f(\pm3) = 0\) | M1 | Look for \(\pm3\) embedded or one of the calculations correct |
| \(-81+27-3c+12=0 \Rightarrow 3c = -42 \Rightarrow c = -14\) | A1* | Proceeds via correct intermediate equation (\(3c=-42\) or \(3c+42=0\)) before reaching \(c=-14\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(3x^3+3x^2-14x+12 = (x+3)(3x^2-6x+4)\) | M1 A1 | M1: attempts to find \(Q(x)\) by factorisation or division — look for first and last terms \(3x^3+3x^2-14x+12=(x+3)(3x^2...x\pm4)\); A1: \((x+3)(3x^2-6x+4)\) seen on same line or states \(Q(x)=(3x^2-6x+4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(b^2-4ac\) for their \((3x^2-6x+4)\) | M1 | Attempts to find roots/discriminant of their quadratic \(3x^2-6x+4=0\) |
| \(b^2-4ac = -12 < 0 \Rightarrow (3x^2-6x+4)\) has no roots and hence \(f(x)=0\) has 1 root \((=-3)\) | A1 | Requires correct \(Q(x)=3x^2-6x+4\), correct calculation, minimal reason why \(Q(x)\) has no solutions, and minimal reason why \(f(x)=0\) has only one solution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sketch: cubic shape with \(x\)-intercept at \(-1\) or \(-9\) (equivalent coordinates) | B1 | Correct shape and position with \(x\) intercept at \(-1\) or \(-9\); graph in quadrants 1, 2 and 3 with minimum in quad 1 and maximum in quad 2 |
| Intercepts of \((-1, 0)\) and \((0, 12)\) only | B1 | Intercepts at \((-1,0)\) and \((0,12)\) ONLY; condone sketch meeting \(x\)-axis at \((-1,0)\) as it would lose first B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sketch of \(y = -f(x)\): correct shape and position | B1 | Graph in quadrants 2, 3 and 4 with minimum in quad 3 and maximum in quad 4 |
| Intercepts of \((-3, 0)\) and \((0, -12)\) only | B1 | Intercepts at \((-3,0)\) and \((0,-12)\) ONLY; condone sketch meeting \(x\)-axis at \((-3,0)\) as it would lose first B1 |
# Question 13:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $f(\pm3) = 0$ | M1 | Look for $\pm3$ embedded or one of the calculations correct |
| $-81+27-3c+12=0 \Rightarrow 3c = -42 \Rightarrow c = -14$ | A1* | Proceeds via correct intermediate equation ($3c=-42$ or $3c+42=0$) before reaching $c=-14$ |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $3x^3+3x^2-14x+12 = (x+3)(3x^2-6x+4)$ | M1 A1 | M1: attempts to find $Q(x)$ by factorisation or division — look for first and last terms $3x^3+3x^2-14x+12=(x+3)(3x^2...x\pm4)$; A1: $(x+3)(3x^2-6x+4)$ seen on same line or states $Q(x)=(3x^2-6x+4)$ |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $b^2-4ac$ for their $(3x^2-6x+4)$ | M1 | Attempts to find roots/discriminant of their quadratic $3x^2-6x+4=0$ |
| $b^2-4ac = -12 < 0 \Rightarrow (3x^2-6x+4)$ has no roots and hence $f(x)=0$ has 1 root $(=-3)$ | A1 | Requires correct $Q(x)=3x^2-6x+4$, correct calculation, minimal reason why $Q(x)$ has no solutions, and minimal reason why $f(x)=0$ has only one solution |
## Part (d)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| Sketch: cubic shape with $x$-intercept at $-1$ or $-9$ (equivalent coordinates) | B1 | Correct shape and position with $x$ intercept at $-1$ or $-9$; graph in quadrants 1, 2 and 3 with minimum in quad 1 and maximum in quad 2 |
| Intercepts of $(-1, 0)$ and $(0, 12)$ only | B1 | Intercepts at $(-1,0)$ and $(0,12)$ ONLY; condone sketch meeting $x$-axis at $(-1,0)$ as it would lose first B1 |
## Part (d)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Sketch of $y = -f(x)$: correct shape and position | B1 | Graph in quadrants 2, 3 and 4 with minimum in quad 3 and maximum in quad 4 |
| Intercepts of $(-3, 0)$ and $(0, -12)$ only | B1 | Intercepts at $(-3,0)$ and $(0,-12)$ ONLY; condone sketch meeting $x$-axis at $(-3,0)$ as it would lose first B1 |13. $\mathrm { f } ( x ) = 3 x ^ { 3 } + 3 x ^ { 2 } + c x + 12$, where $c$ is a constant
Given that $( x + 3 )$ is a factor of $\mathrm { f } ( x )$,
\begin{enumerate}[label=(\alph*)]
\item show that $c = - 14$
\item Write $\mathrm { f } ( x )$ in the form
$$\mathrm { f } ( x ) = ( x + 3 ) \mathrm { Q } ( x )$$
where $\mathrm { Q } ( x )$ is a quadratic function.
\item Use the answer to part (b) to prove that the equation $\mathrm { f } ( x ) = 0$ has only one real solution.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{75d68987-2314-4c8f-8160-24977c5c4e34-32_595_915_1034_518}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.
On separate diagrams sketch the curve with equation
\item \begin{enumerate}[label=(\roman*)]
\item $y = \mathrm { f } ( 3 x )$
\item $y = - \mathrm { f } ( \mathrm { x } )$
On each diagram show clearly the coordinates of the points where the curve crosses the coordinate axes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q13 [10]}}