| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with parameter in coefficient |
| Difficulty | Standard +0.3 This is a standard multi-part Factor and Remainder Theorem question requiring systematic application of well-known techniques: substituting x=3 to get one equation, using the remainder theorem for x=-p to get another, solving simultaneous equations, then factorizing. While it has multiple parts (7 marks total), each step follows directly from textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(3^3 + 3^2(p+3) - 3 + q = 0\) | M1 | Attempts factor theorem setting \(f(3)=0\); score for values embedded leading to equation in \(p\) and \(q\) |
| \(27 + 9p + 27 - 3 + q = 0 \Rightarrow 9p + q = -51\) | A1* | Correct proof, no errors including brackets; must have at least one intermediate stage |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((-p)^3 + (p+3)(-p)^2 - (-p) + q = 9\) | M1 | Attempts remainder theorem setting \(f(\pm p) = 9\); condone sign slips and invisible brackets |
| \(-p^3 + p^3 + 3p^2 + p + q = 9 \Rightarrow 3p^2 + p + q - 9 = 0\) | A1* | Correct proof, no errors; must have at least one intermediate stage |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(3p^2 + p + q - 9 = 0 \Rightarrow 3p^2 + p - 51 - 9p - 9 = 0 \Rightarrow 3p^2 - 8p - 60 = 0\) | M1 | Attempts to use both equations to form a 3TQ in \(p\) (or \(q\)) |
| \(p = 6\) | A1 | Ignore any reference to \(-\frac{10}{3}\) |
| \(q = -51 - 9p = -105\) | A1 | Only |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f(x) = x^3 + 9x^2 - x - 105\), \(f(x) = (x-3)(\ldots x^2 + \ldots x + \ldots)\) | M1 | Uses values of \(p\) and \(q\) with correct strategy (inspection or long division) to obtain quadratic factor |
| \(g(x) = x^2 + 12x + 35\) | A1 | oe e.g. \((x+5)(x+7)\); condone poor notation such as \(f(x) = x^2+12x+35\) |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $3^3 + 3^2(p+3) - 3 + q = 0$ | M1 | Attempts factor theorem setting $f(3)=0$; score for values embedded leading to equation in $p$ and $q$ |
| $27 + 9p + 27 - 3 + q = 0 \Rightarrow 9p + q = -51$ | A1* | Correct proof, no errors including brackets; must have at least one intermediate stage |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(-p)^3 + (p+3)(-p)^2 - (-p) + q = 9$ | M1 | Attempts remainder theorem setting $f(\pm p) = 9$; condone sign slips and invisible brackets |
| $-p^3 + p^3 + 3p^2 + p + q = 9 \Rightarrow 3p^2 + p + q - 9 = 0$ | A1* | Correct proof, no errors; must have at least one intermediate stage |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $3p^2 + p + q - 9 = 0 \Rightarrow 3p^2 + p - 51 - 9p - 9 = 0 \Rightarrow 3p^2 - 8p - 60 = 0$ | M1 | Attempts to use both equations to form a 3TQ in $p$ (or $q$) |
| $p = 6$ | A1 | Ignore any reference to $-\frac{10}{3}$ |
| $q = -51 - 9p = -105$ | A1 | Only |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(x) = x^3 + 9x^2 - x - 105$, $f(x) = (x-3)(\ldots x^2 + \ldots x + \ldots)$ | M1 | Uses values of $p$ and $q$ with correct strategy (inspection or long division) to obtain quadratic factor |
| $g(x) = x^2 + 12x + 35$ | A1 | oe e.g. $(x+5)(x+7)$; condone poor notation such as $f(x) = x^2+12x+35$ |5.
$$f ( x ) = x ^ { 3 } + ( p + 3 ) x ^ { 2 } - x + q$$
where $p$ and $q$ are constants and $p > 0$\\
Given that ( $x - 3$ ) is a factor of $\mathrm { f } ( x )$
\begin{enumerate}[label=(\alph*)]
\item show that
$$9 p + q = - 51$$
Given also that when $\mathrm { f } ( x )$ is divided by ( $x + p$ ) the remainder is 9
\item show that
$$3 p ^ { 2 } + p + q - 9 = 0$$
\item Hence find the value of $p$ and the value of $q$.
\item Hence find a quadratic expression $\mathrm { g } ( x )$ such that
$$f ( x ) = ( x - 3 ) g ( x )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2023 Q5 [9]}}