| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| 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 equations from the given conditions (factor gives p(-1/2)=0, remainder condition gives p(1/2)=1), solve simultaneously for a and b, then apply polynomial division for part (ii). While it involves multiple steps, each step uses standard A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -\frac{1}{2}\), or divide by \((2x + 1)\), and obtain a correct equation, e.g. \(a - 2b + 8 = 0\) | B1 | |
| Substitute \(x = \frac{1}{2}\) and equate to 1, or divide by \((2x - 1)\) and equate constant remainder to 1 | M1 | |
| Obtain a correct equation, e.g. \(a + 2b + 12 = 0\) | A1 | |
| Solve for \(a\) or for \(b\) | M1 | |
| Obtain \(a = -10\) and \(b = -1\) | A1 | [5 marks] |
| (ii) Divide by \(2x^2 - 1\) and reach a quotient of the form \(4x + k\) | M1 | |
| Obtain quotient \(4x - 5\) | A1 | |
| Obtain remainder \(3x - 2\) | A1 | [3 marks total] |
**(i)** Substitute $x = -\frac{1}{2}$, or divide by $(2x + 1)$, and obtain a correct equation, e.g. $a - 2b + 8 = 0$ | B1 |
Substitute $x = \frac{1}{2}$ and equate to 1, or divide by $(2x - 1)$ and equate constant remainder to 1 | M1 |
Obtain a correct equation, e.g. $a + 2b + 12 = 0$ | A1 |
Solve for $a$ or for $b$ | M1 |
Obtain $a = -10$ and $b = -1$ | A1 | [5 marks]
**(ii)** Divide by $2x^2 - 1$ and reach a quotient of the form $4x + k$ | M1 |
Obtain quotient $4x - 5$ | A1 |
Obtain remainder $3x - 2$ | A1 | [3 marks total]
---5 The polynomial $8 x ^ { 3 } + a x ^ { 2 } + b x + 3$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( 2 x + 1 )$ is a factor of $\mathrm { p } ( x )$ and that when $\mathrm { p } ( x )$ is divided by ( $2 x - 1$ ) the remainder is 1 .\\
(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 $2 x ^ { 2 } - 1$.
\hfill \mbox{\textit{CAIE P3 2013 Q5 [8]}}