| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find quotient and remainder by division |
| Difficulty | Easy -1.2 This is a straightforward application of the factor theorem requiring substitution of x=-7 to verify the factor, followed by polynomial division to find the quotient. Both are routine A-level techniques with no problem-solving or insight required, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| 7(a) | Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Factors must be stated again in (b) to collect marks there | B1 | No errors allowed. |
| Answer | Marks |
|---|---|
| 7(b) | Commence division and reach partial quotient of the form 8x2 ± 2x |
| Answer | Marks | Guidance |
|---|---|---|
| = ± 2 or C = –3 | M1 | Condone no visible working. |
| Answer | Marks | Guidance |
|---|---|---|
| Stating (x + 7)(8x2 – 2x – 3) is sufficient | A1 | Division can terminate with 0 or −3x – 21 stated once or twice. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(c) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | M1 | (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0 |
| Answer | Marks |
|---|---|
| Obtain one answer, e.g. 𝜃 = 120° | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (condone 319°) and no others in the interval | A1 | Accept more accurate answers. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0
This is sufficient if no errors seen.
[ – 2744 + 2646 +119 – 21 = 0]
Or complete division of 8x3 + 54x2 – 17x – 21 by x + 7 to get
quotient 8x2 – 2x – 3 and remainder of 0
Or state (x + 7)(8x2 – 2x – 3) is sufficient
Factors must be stated again in (b) to collect marks there | B1 | No errors allowed.
Correct division:
8x2 −2x −3 .
x + 7 8x3 + 54x2 − 17x − 21
8x3 + 56x2 .
− 2x2 −17x
− 2x2 −14x .
− 3x − 21
− 3x − 21
1
--- 7(b) ---
7(b) | Commence division and reach partial quotient of the form 8x2 ± 2x
or
8x3 + 54x2 – 17x – 21 = (x + 7)(Ax2 + Bx + C) and reach A = 8 and B
= ± 2 or C = –3 | M1 | Condone no visible working.
Obtain quotient 8x2 – 2x – 3 with no errors seen
Stating (x + 7)(8x2 – 2x – 3) is sufficient | A1 | Division can terminate with 0 or −3x – 21 stated once or twice.
The working of division and finding quotient may be seen in (a)
but results required here to collect marks.
2
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1
Solve quadratic from (b) to obtain a value for 𝜃 = cos1
2
3
or cos1
4 | M1 | (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0
2 496 1 3
x = cos and .
16 2 4
Obtain one answer, e.g. 𝜃 = 120° | A1
Obtain three further answers, e.g. 𝜃 = 240°, 41.4° and 318.6°
(condone 319°) and no others in the interval | A1 | Accept more accurate answers.
Answers in radians, maximum 2/3.
3
Question | Answer | Marks | GuidanceLet $f(x) = 8x^3 + 54x^2 - 17x - 21$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x + 7$ is a factor of $f(x)$. [1]
\item Find the quotient when $f(x)$ is divided by $x + 7$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q7 [3]}}