| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two factors given |
| Difficulty | Standard +0.3 This is a straightforward application of the factor theorem requiring substitution to find two unknowns, followed by routine factorization and an exponential equation solved by substitution. The multi-part structure and exponential twist elevate it slightly above average, but all techniques are standard A-level procedures with no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State \(-40 + 4a + b = 0\) or equivalent | B1 | |
| State \(-135 + 9a + b = 0\) or equivalent | B1 | |
| Solve a pair of linear simultaneous equations | M1 | |
| Obtain \(a = 19\) and \(b = -36\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identify \(5x - 6\) as a factor | B1 | |
| State \((x+2)(x+3)(5x-6)\) | B1 | |
| State or imply \(5^y = \dfrac{6}{5}\), following a positive value from factorisation | B1\(\sqrt{}\) | |
| Apply logarithms and use power law | M1 | |
| Obtain \(0.113\) only | A1 | [5] |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State $-40 + 4a + b = 0$ or equivalent | B1 | |
| State $-135 + 9a + b = 0$ or equivalent | B1 | |
| Solve a pair of linear simultaneous equations | M1 | |
| Obtain $a = 19$ and $b = -36$ | A1 | [4] |
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identify $5x - 6$ as a factor | B1 | |
| State $(x+2)(x+3)(5x-6)$ | B1 | |
| State or imply $5^y = \dfrac{6}{5}$, following a positive value from factorisation | B1$\sqrt{}$ | |
| Apply logarithms and use power law | M1 | |
| Obtain $0.113$ only | A1 | [5] |
---5 (i) Given that ( $x + 2$ ) and ( $x + 3$ ) are factors of
$$5 x ^ { 3 } + a x ^ { 2 } + b$$
find the values of the constants $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, factorise
$$5 x ^ { 3 } + a x ^ { 2 } + b$$
completely, and hence solve the equation
$$5 ^ { 3 y + 1 } + a \times 5 ^ { 2 y } + b = 0$$
giving any answers correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P2 2014 Q5 [9]}}