| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constant then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring students to substitute x = -1/3, solve for a, then perform polynomial division or factorization. The arithmetic involves fractions but the method is routine and well-practiced at A-level, making it easier than average with no novel problem-solving required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -\frac{1}{3}\), or divide by \(3x + 1\), and obtain a correct equation, e.g. \(-\frac{1}{27}a - \frac{20}{9} - \frac{1}{3} + 3 = 0\) | B1 | |
| Solve for \(a\) an equation obtained by a valid method | M1 | |
| Obtain \(a = 12\) | A1 | [3] |
| (ii) Commence division by \(3x + 1\) reaching a partial quotient \(\frac{1}{3}ax^2 + kx\) | M1 | |
| Obtain quadratic factor \(4x^2 - 8x + 3\) | A1 | |
| Obtain factorisation \((3x + 1)(2x - 1)(2x - 3)\) | A1 | [3] |
**(i)** Substitute $x = -\frac{1}{3}$, or divide by $3x + 1$, and obtain a correct equation, e.g. $-\frac{1}{27}a - \frac{20}{9} - \frac{1}{3} + 3 = 0$ | B1 |
Solve for $a$ an equation obtained by a valid method | M1 |
Obtain $a = 12$ | A1 | [3]
**(ii)** Commence division by $3x + 1$ reaching a partial quotient $\frac{1}{3}ax^2 + kx$ | M1 |
Obtain quadratic factor $4x^2 - 8x + 3$ | A1 |
Obtain factorisation $(3x + 1)(2x - 1)(2x - 3)$ | A1 | [3]
[The M1 is earned if inspection reaches an unknown factor $\frac{1}{3}ax^2 + Bx + C$ and an equation in $B$ and/or $C$, or an unknown factor $Ax^2 + Bx + 3$ and an equation in $A$ and/or $B$, or if two coefficients with the correct moduli are stated without working.]
[If linear factors are found by the factor theorem, give B1B1 for $(2x - 1)$ and $(2x - 3)$, and B1 for the complete factorisation.]
[Synthetic division giving $12x^2 - 24x + 9$ as quadratic factor earns M1A1, but the final factorisation needs $\left(x + \frac{1}{3}\right)$, or equivalent, in order to earn the second A1.]
[SR: If $x = -\frac{1}{3}$ is used in substitution or synthetic division, give the M1 in part (i) but give M0 in part (ii).]The polynomial $ax^3 - 20x^2 + x + 3$, where $a$ is a constant, is denoted by $\text{p}(x)$. It is given that $(3x + 1)$ is a factor of $\text{p}(x)$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $a$. [3]
\item When $a$ has this value, factorise $\text{p}(x)$ completely. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q4 [6]}}