WJEC Further Unit 1 2024 June — Question 2 3 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeGiven factor, find all roots
DifficultyModerate -0.5 This is a straightforward application of polynomial division where the quadratic factor is given. Students divide to find the linear factor, then solve both factors. The quadratic requires completing the square or the quadratic formula for complex roots, making it slightly easier than average since the method is prescribed and involves standard techniques with no problem-solving required.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02g Conjugate pairs: real coefficient polynomials

2. Given that \(x ^ { 2 } + 4 x + 5\) is a factor of \(x ^ { 3 } + x ^ { 2 } - 7 x - 15\), solve the equation \(x ^ { 3 } + x ^ { 2 } - 7 x - 15 = 0\).

Question 2:
AnswerMarks
23
Question 2:
2 | 3
2. Given that $x ^ { 2 } + 4 x + 5$ is a factor of $x ^ { 3 } + x ^ { 2 } - 7 x - 15$, solve the equation $x ^ { 3 } + x ^ { 2 } - 7 x - 15 = 0$.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2024 Q2 [3]}}