CAIE P2 2009 November — Question 5 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2009
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeTwo factors given
DifficultyModerate -0.8 This is a straightforward application of the Factor Theorem requiring substitution of x=-1 and x=2 to create two simultaneous equations, then solving for a and b. Part (ii) involves simple factorization or polynomial division. The question is routine with clear structure and standard techniques, making it easier than average but not trivial since it requires algebraic manipulation of simultaneous equations.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

5 The polynomial \(a x ^ { 3 } + b x ^ { 2 } - 5 x + 2\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). It is given that \(( x + 1 )\) and \(( x - 2 )\) are factors of \(\mathrm { p } ( x )\).
  1. Find the values of \(a\) and \(b\).
  2. When \(a\) and \(b\) have these values, find the other linear factor of \(\mathrm { p } ( x )\).

(i)
AnswerMarks
Substitute \(x = -1\) or \(x = 2\) and equate to zeroM1
Obtain a correct equation, e.g. \(-a + b + 5 + 2 = 0\)A1
Obtain a second correct equation, e.g. \(8a + 4b - 10 + 2 = 0\)A1
Solve for \(a\) or \(b\)M1
Obtain \(a = 3\) and \(b = -4\)A1
Total: [5]
(ii)
AnswerMarks
Substitute for \(a\) and \(b\) and attempt division by \((x + 1)(x - 2)\) or attempt third factor by inspectionM1
Obtain answer \(3x - 1\)A1
Total: [2]
**(i)**
| Substitute $x = -1$ or $x = 2$ and equate to zero | M1 |
| Obtain a correct equation, e.g. $-a + b + 5 + 2 = 0$ | A1 |
| Obtain a second correct equation, e.g. $8a + 4b - 10 + 2 = 0$ | A1 |
| Solve for $a$ or $b$ | M1 |
| Obtain $a = 3$ and $b = -4$ | A1 |

**Total: [5]**

**(ii)**
| Substitute for $a$ and $b$ and attempt division by $(x + 1)(x - 2)$ or attempt third factor by inspection | M1 |
| Obtain answer $3x - 1$ | A1 |

**Total: [2]**

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5 The polynomial $a x ^ { 3 } + b x ^ { 2 } - 5 x + 2$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( x + 1 )$ and $( x - 2 )$ are factors of $\mathrm { p } ( x )$.\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, find the other linear factor of $\mathrm { p } ( x )$.

\hfill \mbox{\textit{CAIE P2 2009 Q5 [7]}}