CAIE P3 2012 June — Question 3 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeSolve p(algebraic transform) = 0
DifficultyStandard +0.3 This is a straightforward multi-part question on the factor theorem. Part (i) requires simple substitution to find a constant. Part (ii)(a) is routine factorization after finding one factor. Part (ii)(b) involves a standard algebraic transformation p(x²)=0, requiring students to solve x²=r for each root r—a common exam technique but slightly above routine drill. Overall slightly easier than average due to clear structure and standard methods.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

3 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = x ^ { 3 } - 3 a x + 4 a$$ where \(a\) is a constant.
  1. Given that \(( x - 2 )\) is a factor of \(\mathrm { p } ( x )\), find the value of \(a\).
  2. When \(a\) has this value,
    1. factorise \(\mathrm { p } ( x )\) completely,
    2. find all the roots of the equation \(\mathrm { p } \left( x ^ { 2 } \right) = 0\).

Question 3:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitute \(x = 2\) and equate to zero, or divide by \(x - 2\) and equate constant remainder to zero, or equivalentM1
Obtain \(a = 4\)A1 [2]
Part (ii)(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Find further (quadratic or linear) factor by division, inspection or factor theorem or equivalentM1
Obtain \(x^2 + 2x - 8\) or \(x + 4\)A1
State \((x-2)^2(x+4)\) or equivalentA1 [3]
Part (ii)(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
State any two of the four (or six) rootsB1\(\checkmark\)
State all roots \((\pm\sqrt{2},\ \pm 2i)\), provided two are purely imaginaryB1\(\checkmark\) [2]
## Question 3:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = 2$ and equate to zero, or divide by $x - 2$ and equate constant remainder to zero, or equivalent | M1 | |
| Obtain $a = 4$ | A1 | [2] |

### Part (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find further (quadratic or linear) factor by division, inspection or factor theorem or equivalent | M1 | |
| Obtain $x^2 + 2x - 8$ or $x + 4$ | A1 | |
| State $(x-2)^2(x+4)$ or equivalent | A1 | [3] |

### Part (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| State any two of the four (or six) roots | B1$\checkmark$ | |
| State all roots $(\pm\sqrt{2},\ \pm 2i)$, provided two are purely imaginary | B1$\checkmark$ | [2] |

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3 The polynomial $\mathrm { p } ( x )$ is defined by

$$\mathrm { p } ( x ) = x ^ { 3 } - 3 a x + 4 a$$

where $a$ is a constant.\\
(i) Given that $( x - 2 )$ is a factor of $\mathrm { p } ( x )$, find the value of $a$.\\
(ii) When $a$ has this value,
\begin{enumerate}[label=(\alph*)]
\item factorise $\mathrm { p } ( x )$ completely,
\item find all the roots of the equation $\mathrm { p } \left( x ^ { 2 } \right) = 0$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2012 Q3 [7]}}