| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two factors given |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring students to substitute x=3 and x=-2 to create two simultaneous equations, solve for a and b, then factorise. It's routine bookwork with clear methodology and no conceptual challenges, making it easier than average but not trivial since it requires careful algebraic manipulation across multiple steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = 3\) or \(x = -2\) and equate to zero | M1 | |
| Obtain a correct equation in any form | A1 | |
| Obtain a second correct equation in any form | A1 | |
| Solve a relevant pair of equations for \(a\) or for \(b\) | M1 | |
| Obtain \(a = 4\) and \(b = -3\) | A1 | [5] |
| (ii) Attempt division by \(x + 2\) (or \(x - 3\)) and obtain partial quotient of \(ax^2 + kx\) | M1 | |
| Obtain linear factors \(4x + 1, x + 2\) and \(x - 3\) | A1 | |
| [If linear factor \(4x + 1\) obtained by remainder theorem or inspection, award B2] | [2] | |
| [If linear factor \(4x + 1\) obtained by division by \(x^2 - x - 6\), award M1 A1] |
| Answer | Marks |
|---|---|
| Attempt to form identity \((x^2 - x - 6)(rx + s) = ax^3 + bx^2 - 25x - 6\) | M1 |
| Attempt to equate like terms | M1 |
| Leads to \(s = 1\) B1, \(r = 4\) A1, \(b = -3\) A1, \(a = 4\) | A1 |
| Obtain linear factors \(4x + 1, x + 2\) and \(x - 3\) | A1 |
(i) Substitute $x = 3$ or $x = -2$ and equate to zero | M1 |
Obtain a correct equation in any form | A1 |
Obtain a second correct equation in any form | A1 |
Solve a relevant pair of equations for $a$ or for $b$ | M1 |
Obtain $a = 4$ and $b = -3$ | A1 | [5]
(ii) Attempt division by $x + 2$ (or $x - 3$) and obtain partial quotient of $ax^2 + kx$ | M1 |
Obtain linear factors $4x + 1, x + 2$ and $x - 3$ | A1 |
[If linear factor $4x + 1$ obtained by remainder theorem or inspection, award B2] | [2] |
[If linear factor $4x + 1$ obtained by division by $x^2 - x - 6$, award M1 A1] | |
**Alternative Method:**
Attempt to form identity $(x^2 - x - 6)(rx + s) = ax^3 + bx^2 - 25x - 6$ | M1 |
Attempt to equate like terms | M1 |
Leads to $s = 1$ B1, $r = 4$ A1, $b = -3$ A1, $a = 4$ | A1 |
Obtain linear factors $4x + 1, x + 2$ and $x - 3$ | A1 |4 (i) The polynomial $a x ^ { 3 } + b x ^ { 2 } - 25 x - 6$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( x - 3 )$ and $( x + 2 )$ are factors of $\mathrm { p } ( x )$. 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 2013 Q4 [7]}}