CAIE P3 2019 November — Question 2 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
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 arithmetic is routine (substituting x = -1/2 and x = -2), and solving the resulting linear system is standard. Below average difficulty as it's a direct textbook-style exercise with no conceptual challenges.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

2 The polynomial \(6 x ^ { 3 } + a x ^ { 2 } + b x - 2\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). It is given that \(( 2 x + 1 )\) is a factor of \(\mathrm { p } ( x )\) and that when \(\mathrm { p } ( x )\) is divided by \(( x + 2 )\) the remainder is - 24 . Find the values of \(a\) and \(b\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Substitute \(x = -\frac{1}{2}\), equate result to zero and obtain a correct equation, e.g. \(-\frac{6}{8} + \frac{1}{4}a - \frac{1}{2}b - 2 = 0\)B1
Substitute \(x = -2\) and equate result to \(-24\)\*M1
Obtain a correct equation, e.g. \(-48 + 4a - 2b - 2 = -24\)A1
Solve for \(a\) or for \(b\)DM1
Obtain \(a = 5\) and \(b = -3\)A1
Total5
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -\frac{1}{2}$, equate result to zero and obtain a correct equation, e.g. $-\frac{6}{8} + \frac{1}{4}a - \frac{1}{2}b - 2 = 0$ | B1 | |
| Substitute $x = -2$ and equate result to $-24$ | \*M1 | |
| Obtain a correct equation, e.g. $-48 + 4a - 2b - 2 = -24$ | A1 | |
| Solve for $a$ or for $b$ | DM1 | |
| Obtain $a = 5$ and $b = -3$ | A1 | |
| **Total** | **5** | |

---
2 The polynomial $6 x ^ { 3 } + a x ^ { 2 } + b x - 2$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( 2 x + 1 )$ is a factor of $\mathrm { p } ( x )$ and that when $\mathrm { p } ( x )$ is divided by $( x + 2 )$ the remainder is - 24 . Find the values of $a$ and $b$.\\

\hfill \mbox{\textit{CAIE P3 2019 Q2 [5]}}