CAIE P2 2010 June — Question 4 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypePolynomial Division by Quadratic Divisor
DifficultyModerate -0.8 This question tests routine polynomial division and direct application of the factor theorem. Part (i) is a standard algebraic long division exercise requiring careful arithmetic but no problem-solving insight. Part (ii) is straightforward substitution to verify f(-1)=0. Both parts are mechanical procedures commonly practiced in textbooks, making this easier than average for A-level.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division

4 The polynomial \(x ^ { 3 } + 3 x ^ { 2 } + 4 x + 2\) is denoted by \(\mathrm { f } ( x )\).
  1. Find the quotient and remainder when \(\mathrm { f } ( x )\) is divided by \(x ^ { 2 } + x - 1\).
  2. Use the factor theorem to show that \(( x + 1 )\) is a factor of \(\mathrm { f } ( x )\).

AnswerMarks Guidance
(i) Commence division by \(x^2 + x - 1\) obtaining quotient of the form \(x + k\)M1
Obtain quotient \(x + 2\)A1
Obtain remainder \(3x + 4\)A1
Identify the quotient and remainder correctlyA1√ [4]
(ii) Substitute \(x = -1\) and evaluate expressionM1
Obtain answer \(0\)A1 [2]
**(i)** Commence division by $x^2 + x - 1$ obtaining quotient of the form $x + k$ | M1 |
Obtain quotient $x + 2$ | A1 |
Obtain remainder $3x + 4$ | A1 |
Identify the quotient and remainder correctly | A1√ | [4]

**(ii)** Substitute $x = -1$ and evaluate expression | M1 |
Obtain answer $0$ | A1 | [2]
4 The polynomial $x ^ { 3 } + 3 x ^ { 2 } + 4 x + 2$ is denoted by $\mathrm { f } ( x )$.\\
(i) Find the quotient and remainder when $\mathrm { f } ( x )$ is divided by $x ^ { 2 } + x - 1$.\\
(ii) Use the factor theorem to show that $( x + 1 )$ is a factor of $\mathrm { f } ( x )$.

\hfill \mbox{\textit{CAIE P2 2010 Q4 [6]}}