| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor and remainder theorems requiring students to set up two simultaneous equations (p(2)=0 and p(1)=4) and solve for a and b, then factorize. It's routine algebraic manipulation with no conceptual challenges beyond standard A-level techniques, making it easier than average but not trivial. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = 2\), equate to zero and state a correct equation, e.g. \(8 + 4a + 2b + 6 = 0\) | B1 | |
| Substitute \(x = 1\) and equate to 4 | M1 | |
| Obtain a correct equation. e.g. \(1 + a + b + 6 = 4\) | A1 | |
| Solve for \(a\) or for \(b\) | M1 | |
| Obtain \(a = -4\) and \(b = 1\) | A1 | [5] |
| (ii) EITHER: Attempt division by \(x - 2\) reaching a partial quotient of \(x^2 + kx\) | M1 | |
| Obtain remainder quadratic factor \(x^2 - 2x - 3\) | A1 | |
| State linear factors \((x - 3)\) and \((x + 1)\) | A1 | |
| OR: Obtain linear factor \((x + 1)\) by inspection | B1 | |
| Obtain factor \((x - 3)\) similarly | B2 | [3] |
**(i)** Substitute $x = 2$, equate to zero and state a correct equation, e.g. $8 + 4a + 2b + 6 = 0$ | B1 |
Substitute $x = 1$ and equate to 4 | M1 |
Obtain a correct equation. e.g. $1 + a + b + 6 = 4$ | A1 |
Solve for $a$ or for $b$ | M1 |
Obtain $a = -4$ and $b = 1$ | A1 | [5]
**(ii)** **EITHER:** Attempt division by $x - 2$ reaching a partial quotient of $x^2 + kx$ | M1 |
Obtain remainder quadratic factor $x^2 - 2x - 3$ | A1 |
State linear factors $(x - 3)$ and $(x + 1)$ | A1 |
**OR:** Obtain linear factor $(x + 1)$ by inspection | B1 |
Obtain factor $(x - 3)$ similarly | B2 | [3]6 The polynomial $x ^ { 3 } + a x ^ { 2 } + b x + 6$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( x - 2 )$ is a factor of $\mathrm { p } ( x )$, and that when $\mathrm { p } ( x )$ is divided by $( x - 1 )$ the remainder is 4 .\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, find the other two linear factors of $\mathrm { p } ( x )$.
\hfill \mbox{\textit{CAIE P2 2009 Q6 [8]}}