| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Remainder condition then further work |
| Difficulty | Moderate -0.3 This is a straightforward application of the Factor Theorem requiring substitution to find a constant, followed by routine factorisation and remainder calculation. While it involves multiple parts, each step uses standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute 2 and equate to zero or divide and equate remainder to zero | M1 | |
| Obtain \(a = 2\) | A1 | [2] |
| (ii) (a) Attempt to find quadratic factor by division, inspection or identity | M1 | |
| Obtain \(2x^2 + x - 3\) | A1 | |
| Conclude \((x - 2)(2x + 3)(x - 1)\) | A1 | [3] |
| (ii) (b) Attempt substitution of \(-1\) or attempt complete division by \(x + 1\) | M1 | |
| Obtain 6 | A1 | [2] |
**(i)** Substitute 2 and equate to zero or divide and equate remainder to zero | M1 |
Obtain $a = 2$ | A1 | [2]
**(ii) (a)** Attempt to find quadratic factor by division, inspection or identity | M1 |
Obtain $2x^2 + x - 3$ | A1 |
Conclude $(x - 2)(2x + 3)(x - 1)$ | A1 | [3]
**(ii) (b)** Attempt substitution of $-1$ or attempt complete division by $x + 1$ | M1 |
Obtain 6 | A1 | [2]3 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = a x ^ { 3 } - 3 x ^ { 2 } - 5 x + a + 4 ,$$
where $a$ is a constant.\\
(i) Given that $( x - 2 )$ is a factor of $\mathrm { p } ( x )$, find the value of $a$.\\
(ii) When $a$ has this value,
\begin{enumerate}[label=(\alph*)]
\item factorise $\mathrm { p } ( x )$ completely,
\item find the remainder when $\mathrm { p } ( x )$ is divided by $( x + 1 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2012 Q3 [7]}}