| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single unknown constant |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution of x=-1 to find a constant, followed by a simple remainder theorem calculation. Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires correct algebraic manipulation. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -1\) and equate to zero | M1 | |
| Obtain answer \(a = 7\) | A1 | [2] |
| (ii) Substitute \(x = -3\) and evaluate expression | M1 | |
| Obtain answer 18 | A1 | [2] |
**(i)** Substitute $x = -1$ and equate to zero | M1 |
Obtain answer $a = 7$ | A1 | [2]
**(ii)** Substitute $x = -3$ and evaluate expression | M1 |
Obtain answer 18 | A1 | [2]3 (i) The polynomial $2 x ^ { 3 } + a x ^ { 2 } - a x - 12$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that $( x + 1 )$ is a factor of $\mathrm { p } ( x )$. Find the value of $a$.\\
(ii) When $a$ has this value, find the remainder when $\mathrm { p } ( x )$ is divided by $( x + 3 )$.
\hfill \mbox{\textit{CAIE P2 2013 Q3 [4]}}