CAIE P3 2021 March — Question 2 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2021
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeSingle polynomial, two remainder/factor conditions
DifficultyModerate -0.8 This is a straightforward application of the factor and remainder theorems requiring students to set up two simultaneous equations (p(-2)=0 and p(-1)=2) and solve for a and b. It's routine algebraic manipulation with no conceptual challenges beyond direct theorem application, making it easier than average but not trivial due to the simultaneous equation solving.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

The polynomial \(ax^3 + 5x^2 - 4x + b\), where \(a\) and \(b\) are constants, is denoted by p\((x)\). It is given that \((x + 2)\) is a factor of p\((x)\) and that when p\((x)\) is divided by \((x + 1)\) the remainder is 2. Find the values of \(a\) and \(b\). [5]

Question 2:
AnswerMarks
2Substitute x = – 2, equate result to zero and obtain a correct equation,
e.g. −8a+20+8+b=0B1
Substitute x = – 1 and equate result to 2M1
Obtain a correct equation, e.g. – a + 5 + 4 + b = 2A1
Solve for a or for bM1
Obtain a = 3 and b = – 4A1
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | Substitute x = – 2, equate result to zero and obtain a correct equation,
e.g. −8a+20+8+b=0 | B1
Substitute x = – 1 and equate result to 2 | M1
Obtain a correct equation, e.g. – a + 5 + 4 + b = 2 | A1
Solve for a or for b | M1
Obtain a = 3 and b = – 4 | A1
5
Question | Answer | Marks | Guidance
The polynomial $ax^3 + 5x^2 - 4x + b$, where $a$ and $b$ are constants, is denoted by p$(x)$. It is given that $(x + 2)$ is a factor of p$(x)$ and that when p$(x)$ is divided by $(x + 1)$ the remainder is 2.

Find the values of $a$ and $b$. [5]

\hfill \mbox{\textit{CAIE P3 2021 Q2 [5]}}