| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| 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)=12) to find a and b, then factorize. It's routine bookwork with clear signposting and standard algebraic manipulation, making it easier than average but not trivial due to the simultaneous equations and cubic factorization steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -2\), equate to zero and obtain a correct equation in any form | B1 | |
| Substitute \(x = -1\) and equate to 12 | M1 | |
| Obtain a correct equation in any form | A1 | |
| Solve a relevant pair of equations for \(a\) or \(b\) | M1 | |
| Obtain \(a = 2\) and \(b = 6\) | A1 | [5] |
| (ii) Attempt division by \(x + 2\) and reach a partial quotient of \(2x^2 - 7x\) | M1 | |
| Obtain quotient \(2x^2 - 7x + 3\) | A1 | |
| Obtain linear factors \(2x - 1\) and \(x - 3\) | A1 | |
| [Condone omission of repetition that \(x + 2\) is a factor.] | ||
| [If linear factors \(2x - 1, x - 3\) obtained by remainder theorem or inspection, award B2 + B1.] | ||
| S.C. M1A1√ if \(a, b\) not both correct | [3] |
**(i)** Substitute $x = -2$, equate to zero and obtain a correct equation in any form | B1 |
Substitute $x = -1$ and equate to 12 | M1 |
Obtain a correct equation in any form | A1 |
Solve a relevant pair of equations for $a$ or $b$ | M1 |
Obtain $a = 2$ and $b = 6$ | A1 | [5] |
**(ii)** Attempt division by $x + 2$ and reach a partial quotient of $2x^2 - 7x$ | M1 |
Obtain quotient $2x^2 - 7x + 3$ | A1 |
Obtain linear factors $2x - 1$ and $x - 3$ | A1 |
[Condone omission of repetition that $x + 2$ is a factor.] | |
[If linear factors $2x - 1, x - 3$ obtained by remainder theorem or inspection, award B2 + B1.] | |
S.C. M1A1√ if $a, b$ not both correct | [3] |7 The polynomial $a x ^ { 3 } - 3 x ^ { 2 } - 11 x + b$, 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 $( x + 1 )$ the remainder is 12 .\\
(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]}}