AQA Further Paper 2 2024 June — Question 4 1 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFactorise polynomial completely
DifficultyModerate -0.8 This is a straightforward recall question testing the fundamental property that complex roots of polynomials with real coefficients come in conjugate pairs. Since 5i is a root, -5i must also be a root, giving factor (x-5i)(x+5i) = x²+25. Requires only one conceptual step with no calculation or problem-solving.
Spec4.02g Conjugate pairs: real coefficient polynomials

The function f is a quartic function with real coefficients. The complex number \(5i\) is a root of the equation \(f(x) = 0\) Which one of the following must be a factor of \(f(x)\)? Circle your answer. [1 mark] \((x^2 - 25)\) \quad\quad \((x^2 - 5)\) \quad\quad \((x^2 + 5)\) \quad\quad \((x^2 + 25)\)

Question 4:
AnswerMarks Guidance
4Circles 4th answer 2.2a
Question total1
QMarking instructions AO
Question 4:
4 | Circles 4th answer | 2.2a | B1 | (x2 + 25)
Question total | 1
Q | Marking instructions | AO | Marks | Typical solution
The function f is a quartic function with real coefficients.

The complex number $5i$ is a root of the equation $f(x) = 0$

Which one of the following must be a factor of $f(x)$?

Circle your answer.
[1 mark]

$(x^2 - 25)$ \quad\quad $(x^2 - 5)$ \quad\quad $(x^2 + 5)$ \quad\quad $(x^2 + 25)$

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q4 [1]}}