CAIE P3 2009 November — Question 5 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2009
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeMultiple unknowns with derivative condition
DifficultyStandard +0.3 This is a straightforward application of the factor theorem combined with differentiation. Students substitute x = -2 into both p(x) and p'(x) to create two simultaneous equations in a and b, then solve. The complete factorisation follows directly. While it requires coordination of multiple techniques, each step is routine and the problem structure is clearly signposted, making it slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums

5 The polynomial \(2 x ^ { 3 } + a x ^ { 2 } + b x - 4\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). The result of differentiating \(\mathrm { p } ( x )\) with respect to \(x\) is denoted by \(\mathrm { p } ^ { \prime } ( x )\). It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\) and of \(\mathrm { p } ^ { \prime } ( x )\).
  1. Find the values of \(a\) and \(b\).
  2. When \(a\) and \(b\) have these values, factorise \(\mathrm { p } ( x )\) completely.

AnswerMarks Guidance
(i) Substitute \(x = -2\), equate to zero and state a correct equation, e.g. \(-16 + 4a - 2b - 4 = 0\)B1
Differentiate p(x), substitute \(x = -2\) and equate to zeroM1
Obtain a correct equation, e.g. \(24 - 4a + b = 0\)A1
Solve for \(a\) or for \(b\)M1
Obtain \(a = 7\) and \(b = 4\)A1 [5]
(ii) EITHER: State or imply \((x + 2)^2\) is a factorB1
Attempt division by \((x + 2)^2\) reaching a quotient \(2x + k\) or use inspection with unknown factor \(cx + d\) reaching \(c = 2\) or \(d = -1\)M1
Obtain factorisation \((x + 2)^2(2x - 1)\)A1
OR: Attempt division by \((x + 2)\)M1
Obtain quadratic factor \(2x^2 + 3x - 2\)A1
Obtain factorisation \((x + 2)(x + 2)(2x - 1)\)A1 [3]
[The M1 is earned if division reaches a partial quotient of \(2x^2 + kx\), or if inspection has an unknown factor of \(2x^2 + cx + f\) and an equation in e and/or f, or if two coefficients with the correct moduli are stated without working.]
**(i)** Substitute $x = -2$, equate to zero and state a correct equation, e.g. $-16 + 4a - 2b - 4 = 0$ | B1 |
Differentiate p(x), substitute $x = -2$ and equate to zero | M1 |
Obtain a correct equation, e.g. $24 - 4a + b = 0$ | A1 |
Solve for $a$ or for $b$ | M1 |
Obtain $a = 7$ and $b = 4$ | A1 | [5]

**(ii)** EITHER: State or imply $(x + 2)^2$ is a factor | B1 |
Attempt division by $(x + 2)^2$ reaching a quotient $2x + k$ or use inspection with unknown factor $cx + d$ reaching $c = 2$ or $d = -1$ | M1 |
Obtain factorisation $(x + 2)^2(2x - 1)$ | A1 |
OR: Attempt division by $(x + 2)$ | M1 |
Obtain quadratic factor $2x^2 + 3x - 2$ | A1 |
Obtain factorisation $(x + 2)(x + 2)(2x - 1)$ | A1 | [3]

[The M1 is earned if division reaches a partial quotient of $2x^2 + kx$, or if inspection has an unknown factor of $2x^2 + cx + f$ and an equation in e and/or f, or if two coefficients with the correct moduli are stated without working.]
5 The polynomial $2 x ^ { 3 } + a x ^ { 2 } + b x - 4$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. The result of differentiating $\mathrm { p } ( x )$ with respect to $x$ is denoted by $\mathrm { p } ^ { \prime } ( x )$. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ and of $\mathrm { p } ^ { \prime } ( x )$.\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, factorise $\mathrm { p } ( x )$ completely.

\hfill \mbox{\textit{CAIE P3 2009 Q5 [8]}}