AQA Further AS Paper 1 2018 June — Question 8 5 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyStandard +0.8 This is a Further Maths complex numbers question requiring knowledge that complex roots come in conjugate pairs for real coefficients, followed by finding the third root via sum of roots and then m via Vieta's formulas or substitution. While systematic, it requires multiple connected steps and careful algebraic manipulation beyond standard A-level, placing it moderately above average difficulty.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots
\(2 - 3i\) is one root of the equation $$z^3 + mz + 52 = 0$$ where \(m\) is real.
  1. Find the other roots. [3 marks]
  2. Determine the value of \(m\). [2 marks]
Question 8:
AnswerMarks Guidance
8(a)Correctly identifies the complex conjugate as a root. 1.1b
2+3i
sum of roots
𝑏𝑏
= βˆ’π‘Žπ‘Ž
∴ 𝛼𝛼+2+3i+2βˆ’3i = 0
𝛼𝛼+4 = 0
Forms one of the following equations (or their equivalents)
or
or with an equation to find .
𝛼𝛼+𝛽𝛽+2βˆ’3𝑖𝑖 = 0 𝛼𝛼𝛽𝛽(2βˆ’3𝑖𝑖)= Β±52
Or forms an equation to find and then solves the cubic for their value of .
AnswerMarks Guidance
𝛼𝛼𝛽𝛽+𝛽𝛽(2βˆ’3𝑖𝑖)+(2βˆ’3𝑖𝑖)𝛽𝛽 = Β±π‘šπ‘š π‘šπ‘š1.1a M1
π‘šπ‘š π‘šπ‘š
AnswerMarks Guidance
Finds correct third root1.1b A1
𝛼𝛼 2n=d cβˆ’o4mplex root is
2+3i
product of roots
𝑑𝑑
= βˆ’π‘Žπ‘Ž
∴ 𝛼𝛼(2+3i)(2βˆ’3 i)= βˆ’52
2
𝛼𝛼(4βˆ’9𝑖𝑖 )= βˆ’52
βˆ’52
𝛼𝛼 =
13
AnswerMarks Guidance
QMarking instructions AO
AnswerMarks
8(b)Forms a correct equation to find .
May be seen in part (a).
AnswerMarks Guidance
π‘šπ‘š1.1a M1
∴ (βˆ’4) +π‘šπ‘š(βˆ’4)+52 = 0
βˆ’64βˆ’4π‘šπ‘š+52 = 0
βˆ’12 = 4π‘šπ‘š
Finds correct value of .
AnswerMarks Guidance
π‘šπ‘š1.1b A1
π‘šπ‘š = βˆ’3
𝑐𝑐
�𝛼𝛼𝛽𝛽 =
π‘Žπ‘Ž
∴ (2βˆ’3𝑖𝑖)(2+3𝑖𝑖)+(βˆ’4)(2βˆ’3𝑖𝑖)
+(βˆ’4)( 2+3𝑖𝑖)= π‘šπ‘š
4+9βˆ’8+12π‘–π‘–βˆ’8βˆ’12𝑖𝑖 = π‘šπ‘š
AnswerMarks Guidance
Total5 π‘šπ‘š = βˆ’3
QMarking instructions AO 
Question 8:
--- 8(a) ---
8(a) | Correctly identifies the complex conjugate as a root. | 1.1b | B1 | 2nd complex root is
2+3i
sum of roots
𝑏𝑏
= βˆ’π‘Žπ‘Ž
∴ 𝛼𝛼+2+3i+2βˆ’3i = 0
𝛼𝛼+4 = 0
Forms one of the following equations (or their equivalents)
or
or with an equation to find .
𝛼𝛼+𝛽𝛽+2βˆ’3𝑖𝑖 = 0 𝛼𝛼𝛽𝛽(2βˆ’3𝑖𝑖)= Β±52
Or forms an equation to find and then solves the cubic for their value of .
𝛼𝛼𝛽𝛽+𝛽𝛽(2βˆ’3𝑖𝑖)+(2βˆ’3𝑖𝑖)𝛽𝛽 = Β±π‘šπ‘š π‘šπ‘š | 1.1a | M1
π‘šπ‘š π‘šπ‘š
Finds correct third root | 1.1b | A1
𝛼𝛼 2n=d cβˆ’o4mplex root is
2+3i
product of roots
𝑑𝑑
= βˆ’π‘Žπ‘Ž
∴ 𝛼𝛼(2+3i)(2βˆ’3 i)= βˆ’52
2
𝛼𝛼(4βˆ’9𝑖𝑖 )= βˆ’52
βˆ’52
𝛼𝛼 =
13
Q | Marking instructions | AO | Mark | Typical solution
--- 8(b) ---
8(b) | Forms a correct equation to find .
May be seen in part (a).
π‘šπ‘š | 1.1a | M1 | 3
∴ (βˆ’4) +π‘šπ‘š(βˆ’4)+52 = 0
βˆ’64βˆ’4π‘šπ‘š+52 = 0
βˆ’12 = 4π‘šπ‘š
Finds correct value of .
π‘šπ‘š | 1.1b | A1
π‘šπ‘š = βˆ’3
𝑐𝑐
�𝛼𝛼𝛽𝛽 =
π‘Žπ‘Ž
∴ (2βˆ’3𝑖𝑖)(2+3𝑖𝑖)+(βˆ’4)(2βˆ’3𝑖𝑖)
+(βˆ’4)( 2+3𝑖𝑖)= π‘šπ‘š
4+9βˆ’8+12π‘–π‘–βˆ’8βˆ’12𝑖𝑖 = π‘šπ‘š
Total | 5 | π‘šπ‘š = βˆ’3
Q | Marking instructions | AO | Mark | Typical solution
$2 - 3i$ is one root of the equation
$$z^3 + mz + 52 = 0$$
where $m$ is real.

\begin{enumerate}[label=(\alph*)]
\item Find the other roots.
[3 marks]

\item Determine the value of $m$.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q8 [5]}}
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