| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.3 This is a straightforward application of the Factor and Remainder Theorems requiring students to set up two simultaneous equations from given conditions (p(2)=0 and p(-2)=-20), solve for constants a and b, then apply polynomial division. Part (ii) adds a minor twist using x²-4=(x-2)(x+2), but the solution follows directly from part (i). Standard textbook exercise with clear method and 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 |
|---|---|---|
| Substitute \(x = 2\), equate to zero, and state a correct equation, e.g. \(16 - 12 + 2a + b = 0\) | B1 | |
| Substitute \(x = -2\) and equate to –20 | M1 | |
| Obtain a correct equation, e.g. \(-16 - 12 - 2a + b = -20\) | A1 | |
| Solve for \(a\) or for \(b\) | M1 | |
| Obtain \(a = -3\) and \(b = 2\) | A1 | [5 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt division by \(x^2 - 4\) reaching a partial quotient of \(2x - 3\), or a similar stage by inspection | B1 | |
| Obtain remainder \(5x - 10\) | B1∨ + B1∨ | [3 marks] |
**(i)**
Substitute $x = 2$, equate to zero, and state a correct equation, e.g. $16 - 12 + 2a + b = 0$ | B1 | |
Substitute $x = -2$ and equate to –20 | M1 | |
Obtain a correct equation, e.g. $-16 - 12 - 2a + b = -20$ | A1 | |
Solve for $a$ or for $b$ | M1 | |
Obtain $a = -3$ and $b = 2$ | A1 | [5 marks] |
**(ii)**
Attempt division by $x^2 - 4$ reaching a partial quotient of $2x - 3$, or a similar stage by inspection | B1 | |
Obtain remainder $5x - 10$ | B1∨ + B1∨ | [3 marks] |4 The polynomial $2 x ^ { 3 } - 3 x ^ { 2 } + a x + b$, 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 $( x + 2 )$ the remainder is - 20 .\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, find the remainder when $\mathrm { p } ( x )$ is divided by ( $x ^ { 2 } - 4$ ).
\hfill \mbox{\textit{CAIE P2 2007 Q4 [8]}}