CAIE P2 2023 November — Question 1 2 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2023
SessionNovember
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeNon-zero remainder condition
DifficultyEasy -1.2 This is a straightforward application of the Remainder Theorem requiring substitution of x = -2 and solving a simple linear equation. It's a single-step problem testing basic recall of the theorem with minimal algebraic manipulation, making it easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

1 When the polynomial $$a x ^ { 3 } + 4 a x ^ { 2 } - 7 x - 5$$ is divided by \(( x + 2 )\), the remainder is 33 .
Find the value of the constant \(a\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Substitute \(x = -2\) and equate to 33M1 OE (long or synthetic division). Note: Long division and synthetic division give a remainder of \(8a + 14 - 5\). Allow one sign error for M1.
Obtain \(-8a + 16a + 14 - 5 = 33\) and hence \(a = 3\)A1
Total: 2
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -2$ and equate to 33 | M1 | OE (long or synthetic division). Note: Long division and synthetic division give a remainder of $8a + 14 - 5$. Allow one sign error for M1. |
| Obtain $-8a + 16a + 14 - 5 = 33$ and hence $a = 3$ | A1 | |
| | **Total: 2** | |

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1 When the polynomial

$$a x ^ { 3 } + 4 a x ^ { 2 } - 7 x - 5$$

is divided by $( x + 2 )$, the remainder is 33 .\\
Find the value of the constant $a$.\\

\hfill \mbox{\textit{CAIE P2 2023 Q1 [2]}}