WJEC Further Unit 1 2019 June — Question 5 6 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeFactor theorem and finding roots
DifficultyModerate -0.5 This is a straightforward application of the factor theorem where two roots are given. Students factor out (2x+1)(x+3), perform polynomial division to find the quadratic factor, then solve it. While it's a Further Maths question, it's a routine multi-step exercise requiring only standard techniques with no novel insight, making it slightly easier than average.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
5. Given that \(x = - \frac { 1 } { 2 }\) and \(x = - 3\) are two roots of the equation $$2 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 23 x + 15 = 0$$ find the remaining roots.
Method 1:
AnswerMarks Guidance
\(\left(x + \frac{7}{2}\right)(x + 3)\) leading to \((2x + 1)(x + 3)\). \(2x^2 + 7x + 3\) is the quadratic factor. \(2x^4 - x^3 - 15x^2 + 23x + 15 = 0\), \((2x^2 + 7x + 3)(x^2 - 4x + 5) = 0\)M1, A1, M1A1 Allow \(x^2 + \frac{7}{2}x + \frac{7}{2}\); \((x^2 - 8x + 10) / 2x^2 - 8x + 10\); \(2x^2 - 8x + 10 = 0\)
Solving \(x^2 - 4x + 5 = 0\): \(x = \frac{4\pm\sqrt{16-20}}{2}\) OR \((x-2)^2 = -1\), \(x = 2 + i\) or \(x = 2 - i\)M1, A1 cao; Award if previous M1
Method 2:
AnswerMarks Guidance
Use of roots of polynomials to give: \(\alpha + \beta - \frac{1}{2} - 3 = \frac{1}{2} \rightarrow \alpha + \beta = 4\), \(\alpha \times \beta \times (-\frac{1}{2}) \times (-3) = \frac{15}{2} \rightarrow \alpha\beta = 5\)(M1), (A1), (A1) At least 2 equations
\(\therefore x^2 - 4x + 5 = 0\), Solving: \(x = \frac{4\pm\sqrt{16-20}}{2}\) OR \((x-2)^2 = -1\), \(x = 2 + i\) or \(x = 2 - i\)(A1), (M1), (A1) cao; Award if previous M1 
**Method 1:**
$\left(x + \frac{7}{2}\right)(x + 3)$ leading to $(2x + 1)(x + 3)$. $2x^2 + 7x + 3$ is the quadratic factor. $2x^4 - x^3 - 15x^2 + 23x + 15 = 0$, $(2x^2 + 7x + 3)(x^2 - 4x + 5) = 0$ | M1, A1, M1A1 | Allow $x^2 + \frac{7}{2}x + \frac{7}{2}$; $(x^2 - 8x + 10) / 2x^2 - 8x + 10$; $2x^2 - 8x + 10 = 0$

Solving $x^2 - 4x + 5 = 0$: $x = \frac{4\pm\sqrt{16-20}}{2}$ OR $(x-2)^2 = -1$, $x = 2 + i$ or $x = 2 - i$ | M1, A1 | cao; Award if previous M1

**Method 2:**
Use of roots of polynomials to give: $\alpha + \beta - \frac{1}{2} - 3 = \frac{1}{2} \rightarrow \alpha + \beta = 4$, $\alpha \times \beta \times (-\frac{1}{2}) \times (-3) = \frac{15}{2} \rightarrow \alpha\beta = 5$ | (M1), (A1), (A1) | At least 2 equations

$\therefore x^2 - 4x + 5 = 0$, Solving: $x = \frac{4\pm\sqrt{16-20}}{2}$ OR $(x-2)^2 = -1$, $x = 2 + i$ or $x = 2 - i$ | (A1), (M1), (A1) | cao; Award if previous M1

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5. Given that $x = - \frac { 1 } { 2 }$ and $x = - 3$ are two roots of the equation

$$2 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 23 x + 15 = 0$$

find the remaining roots.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q5 [6]}}
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