CAIE P2 2020 November — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeOne factor, one non-zero remainder
DifficultyModerate -0.8 This is a straightforward application of the factor and remainder theorems requiring two substitutions to form simultaneous equations. The algebra is routine with no conceptual challenges—easier than average A-level content.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

2 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + 16$$ where \(a\) and \(b\) are constants. It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\) and that the remainder is 72 when \(\mathrm { p } ( x )\) is divided by \(( x - 2 )\). Find the values of \(a\) and \(b\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Substitute \(x = -2\) and equate to zero\*M1
Substitute \(x = 2\) and equate to 72\*M1
Obtain \(4a - 2b + 8 = 0\) and \(4a + 2b - 48 = 0\) or equivalentsA1
Solve a pair of relevant linear simultaneous equationsDM1 Dependent on at least one M mark
Obtain \(a = 5\), \(b = 14\)A1
Total5
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -2$ and equate to zero | \*M1 | |
| Substitute $x = 2$ and equate to 72 | \*M1 | |
| Obtain $4a - 2b + 8 = 0$ and $4a + 2b - 48 = 0$ or equivalents | A1 | |
| Solve a pair of relevant linear simultaneous equations | DM1 | Dependent on at least one M mark |
| Obtain $a = 5$, $b = 14$ | A1 | |
| **Total** | **5** | |
2 The polynomial $\mathrm { p } ( x )$ is defined by

$$\mathrm { p } ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + 16$$

where $a$ and $b$ are constants. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ and that the remainder is 72 when $\mathrm { p } ( x )$ is divided by $( x - 2 )$.

Find the values of $a$ and $b$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q2 [5]}}