| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with equal remainders |
| Difficulty | Standard +0.3 This is a straightforward application of the Remainder Theorem requiring substitution of x=1 and x=-1 to form equations, then using the factor theorem with x=2/3. The algebra is routine with no conceptual challenges—slightly easier than average as it's a standard multi-part question with clear signposting and mechanical steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3(1)^3 + A(1)^2 + B(1) - 10 = k\) or \(3(-1)^3 + A(-1)^2 + B(-1) - 10 = -10k\) | M1 | Attempts \(f(\pm 1) = k\) or \(f(\pm 1) = -10k\) |
| \(A + B - 7 = k,\ A - B - 13 = -10k \Rightarrow -10A - 10B + 70 = A - B - 13\) | M1 | Uses \(f(\pm 1) = k\) and \(f(\mp 1) = -10k\) to eliminate \(k\) and obtain equation in \(A\) and \(B\) only |
| \(11A + 9B = 83\) | A1* | Correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\left(\frac{2}{3}\right)^3 + A\left(\frac{2}{3}\right)^2 + B\left(\frac{2}{3}\right) - 10 = 0\) | M1 | Attempts \(f\!\left(\frac{2}{3}\right) = 0\) |
| \(11A + 9B = 83,\ 12A + 18B = 246 \Rightarrow A = \ldots,\ B = \ldots\) | M1 | Solves \(11A + 9B = 83\) simultaneously with their equation in \(A\) and \(B\) |
| \(A = -8,\ B = 19\) | A1 | Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (3x-2)(x^2 + \ldots x + \ldots)\) | M1 | Uses any appropriate method e.g. long division/inspection to obtain \(x^2 + px + q\) where \(p\) and \(q\) are non-zero |
| \(g(x) = x^2 - 2x + 5\) | A1 | Correct expression |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3(1)^3 + A(1)^2 + B(1) - 10 = k$ **or** $3(-1)^3 + A(-1)^2 + B(-1) - 10 = -10k$ | M1 | Attempts $f(\pm 1) = k$ **or** $f(\pm 1) = -10k$ |
| $A + B - 7 = k,\ A - B - 13 = -10k \Rightarrow -10A - 10B + 70 = A - B - 13$ | M1 | Uses $f(\pm 1) = k$ **and** $f(\mp 1) = -10k$ to eliminate $k$ and obtain equation in $A$ and $B$ only |
| $11A + 9B = 83$ | A1* | Correct proof with no errors |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\left(\frac{2}{3}\right)^3 + A\left(\frac{2}{3}\right)^2 + B\left(\frac{2}{3}\right) - 10 = 0$ | M1 | Attempts $f\!\left(\frac{2}{3}\right) = 0$ |
| $11A + 9B = 83,\ 12A + 18B = 246 \Rightarrow A = \ldots,\ B = \ldots$ | M1 | Solves $11A + 9B = 83$ simultaneously with their equation in $A$ and $B$ |
| $A = -8,\ B = 19$ | A1 | Correct values |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (3x-2)(x^2 + \ldots x + \ldots)$ | M1 | Uses any appropriate method e.g. long division/inspection to obtain $x^2 + px + q$ where $p$ and $q$ are non-zero |
| $g(x) = x^2 - 2x + 5$ | A1 | Correct expression |
**Total: 8 marks**
---5.
$$f ( x ) = 3 x ^ { 3 } + A x ^ { 2 } + B x - 10$$
where $A$ and $B$ are integers.\\
Given that
\begin{itemize}
\item when $\mathrm { f } ( x )$ is divided by $( x - 1 )$ the remainder is $k$
\item when $\mathrm { f } ( x )$ is divided by $( x + 1 )$ the remainder is $- 10 k$
\item $k$ is a constant
\begin{enumerate}[label=(\alph*)]
\item show that
\end{itemize}
$$11 A + 9 B = 83$$
Given also that $( 3 x - 2 )$ is a factor of $\mathrm { f } ( x )$,
\item find the value of $A$ and the value of $B$.
\item Hence find the quadratic expression $\mathrm { g } ( x )$ such that
$$f ( x ) = ( 3 x - 2 ) g ( x )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2022 Q5 [8]}}