| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | November |
| 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 standard Factor and Remainder Theorem question requiring systematic application of two conditions to find two unknowns, followed by routine factorization. While it involves multiple steps and solving simultaneous equations, it follows a well-practiced template with no conceptual surprises, making it slightly easier than average for A-level. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (i) Substitute \(x = -1\), equate to zero and simplify at least as far as \(-8 + a - b - 1 = 0\) | B1 |
| Substitute \(x = -\frac{1}{2}\) and equate the result to 1 | M1 |
| Obtain a correct equation in any form, e.g. \(-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1\) | A1 |
| Solve for \(a\) or for \(b\) | M1 |
| Obtain \(a = 6\) and \(b = -3\) | A1 |
| [5] | |
| (ii) Commence division by \((x + 1)\) reaching a partial quotient \(8x^2 + kx\) | M1 |
| Obtain quadratic factor \(8x^2 - 2x - 1\) | A1 |
| Obtain factorisation \((x + 1)(4x + 1)(2x - 1)\) | A1 |
| [3] |
**(i)** Substitute $x = -1$, equate to zero and simplify at least as far as $-8 + a - b - 1 = 0$ | B1 |
Substitute $x = -\frac{1}{2}$ and equate the result to 1 | M1 |
Obtain a correct equation in any form, e.g. $-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1$ | A1 |
Solve for $a$ or for $b$ | M1 |
Obtain $a = 6$ and $b = -3$ | A1 |
| [5] |
**(ii)** Commence division by $(x + 1)$ reaching a partial quotient $8x^2 + kx$ | M1 |
Obtain quadratic factor $8x^2 - 2x - 1$ | A1 |
Obtain factorisation $(x + 1)(4x + 1)(2x - 1)$ | A1 |
| [3] |
[The M1 is earned if inspection reaches an unknown factor $8x^2 + Bx + C$ and an equation in $B$ and/or $C$, or an unknown factor $Ax^2 + Bx - 1$ and an equation in $A$ and/or $B$.]
[If linear factors are found by the factor theorem, give B1B1 for $(2x - 1)$ and $(4x + 1)$, and B1 for the complete factorisation.]6 The polynomial $8 x ^ { 3 } + a x ^ { 2 } + b x - 1$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( x + 1 )$ 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, factorise $\mathrm { p } ( x )$ completely.
\hfill \mbox{\textit{CAIE P3 2015 Q6 [8]}}