| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants using remainder theorem |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem and factor theorem with clear scaffolding. Part (a) requires immediate recognition that the remainder is 21 (already in the correct form), part (b) is a simple substitution of x=1/2 and solving a linear equation, and parts (c)(i)-(ii) follow mechanically once k is found. The question requires only direct recall and routine algebraic manipulation with no problem-solving insight needed. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(21\) | B1 | States 21 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(f\!\left(\pm\frac{1}{2}\right)=0\rightarrow\) equation in \(k\) | M1 | The \("=0"\) may be implied. If division used, look for remainder after dividing by \((2x-1)\) equal to zero |
| \(\left(\frac{1}{2}-2\right)\!\left(2\times\frac{1}{4}+5\times\frac{1}{2}+k\right)+21=0 \Rightarrow 3+k=14\Rightarrow k=11\) | A1* | Solves correct equation in \(k\) showing at least one correct intermediate line. Condone missing brackets if clearly recovered |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x)=(x-2)(2x^2+5x+11)+21=2x^3+x^2+x-1\) | M1 | Attempts to multiply out \((x-2)(2x^2+5x+11)+21\) to achieve a cubic |
| \(=(2x-1)(x^2+x+1)\) | dM1, A1 | dM1 attempts to divide/factor out \((2x-1)\) or \(\left(x-\frac{1}{2}\right)\); A1 for \((2x-1)(x^2+x+1)\) from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts to find number of roots of \((x^2+x+1)\) | M1 | Must use correct factor from cubic; attempt any of: \(b^2-4ac\), quadratic formula, completing the square, or complex roots from calculator |
| States \((x^2+x+1)\) has no real roots with reason | A1 | Requires: correct factorisation, correct reason why \((x^2+x+1)=0\) has no roots e.g. \(b^2-4ac=1-4=-3<0\), AND concludes \(f(x)=0\) has 1 real root at \(x=\frac{1}{2}\) |
| Concludes \(f(x)=0\) has 1 root at \(x=\frac{1}{2}\) | Do not accept "no roots" with no supporting evidence |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $21$ | B1 | States 21 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $f\!\left(\pm\frac{1}{2}\right)=0\rightarrow$ equation in $k$ | M1 | The $"=0"$ may be implied. If division used, look for remainder after dividing by $(2x-1)$ equal to zero |
| $\left(\frac{1}{2}-2\right)\!\left(2\times\frac{1}{4}+5\times\frac{1}{2}+k\right)+21=0 \Rightarrow 3+k=14\Rightarrow k=11$ | A1* | Solves correct equation in $k$ showing at least one correct intermediate line. Condone missing brackets if clearly recovered |
## Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x)=(x-2)(2x^2+5x+11)+21=2x^3+x^2+x-1$ | M1 | Attempts to multiply out $(x-2)(2x^2+5x+11)+21$ to achieve a cubic |
| $=(2x-1)(x^2+x+1)$ | dM1, A1 | dM1 attempts to divide/factor out $(2x-1)$ or $\left(x-\frac{1}{2}\right)$; A1 for $(2x-1)(x^2+x+1)$ from correct work |
## Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to find number of roots of $(x^2+x+1)$ | M1 | Must use correct factor from cubic; attempt any of: $b^2-4ac$, quadratic formula, completing the square, or complex roots from calculator |
| States $(x^2+x+1)$ has no real roots with reason | A1 | Requires: correct factorisation, correct reason why $(x^2+x+1)=0$ has no roots e.g. $b^2-4ac=1-4=-3<0$, AND concludes $f(x)=0$ has 1 real root at $x=\frac{1}{2}$ |
| Concludes $f(x)=0$ has 1 root at $x=\frac{1}{2}$ | | Do not accept "no roots" with no supporting evidence |4.
$$f ( x ) = ( x - 2 ) \left( 2 x ^ { 2 } + 5 x + k \right) + 21$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item State the remainder when $\mathrm { f } ( x )$ is divided by ( $x - 2$ )
Given that ( $2 x - 1$ ) is a factor of $\mathrm { f } ( x )$
\item show that $k = 11$
\item Hence
\begin{enumerate}[label=(\roman*)]
\item fully factorise $\mathrm { f } ( x )$,
\item find the number of real solutions of the equation
$$\mathrm { f } ( x ) = 0$$
giving a reason for your answer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2024 Q4 [8]}}