| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| 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 (using p(-2)=0 and p(-1)=24) and solve for a and b, followed by routine factorisation. The algebraic manipulation is standard with no conceptual challenges, making it slightly easier than average but not trivial due to the multi-step nature and need for careful arithmetic. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -2\) and equate to zero | M1 | |
| Substitute \(x = -1\) and equate to \(24\) | M1 | |
| Obtain \(4a - 2b = 38\) and \(a - b = 20\) or equivalents | A1 | |
| Attempt solution of two linear simultaneous equations (dependent on M1 M1) | M1 | |
| Obtain \(a = -1\) and \(b = -21\) | A1 | [5] |
| (ii) Attempt to find quadratic factor by division, inspection or use of identity | M1 | |
| Obtain \(6x^2 - 13x + 5\) | A1√ | |
| Conclude \((x + 2)(2x - 1)(3x - 5)\) | A1 | [3] |
**(i)** Substitute $x = -2$ and equate to zero | M1 |
Substitute $x = -1$ and equate to $24$ | M1 |
Obtain $4a - 2b = 38$ and $a - b = 20$ or equivalents | A1 |
Attempt solution of two linear simultaneous equations (dependent on M1 M1) | M1 |
Obtain $a = -1$ and $b = -21$ | A1 | [5]
**(ii)** Attempt to find quadratic factor by division, inspection or use of identity | M1 |
Obtain $6x^2 - 13x + 5$ | A1√ |
Conclude $(x + 2)(2x - 1)(3x - 5)$ | A1 | [3]7 The cubic polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + a x ^ { 2 } + b x + 10$$
where $a$ and $b$ are constants. 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 24 .\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, factorise $\mathrm { p } ( x )$ completely.
\hfill \mbox{\textit{CAIE P2 2011 Q7 [8]}}