| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Finding specific root values |
| Difficulty | Standard +0.3 Part (a) requires recognizing that αβγ=0 means one root is zero, then solving a quadratic using Vieta's formulas—straightforward algebra. Part (b) involves a standard transformation of roots (scaling by 3) using either substitution or Vieta's formulas again. This is a routine Further Maths question testing fundamental polynomial theory with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Because \(\alpha\beta\gamma = 0\), put \(\alpha = 0\). Then, \(\beta + \gamma = -9\) and \(\beta\gamma = 20\). Attempt to solve, \(\beta = -4\) and \(\gamma = -5\) | B1 B1 M1 A1 | Accept solutions where variables are interchanged throughout |
| Answer | Marks |
|---|---|
| Answer: Because \(\alpha\beta\gamma = 0\), either \(\alpha = 0\) or \(\beta = 0\) or \(\gamma = 0\) | (B1) |
| Answer: \(ax^3 + bx^2 + cx + d = 0\) becomes \(x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = 0\) | (B1) |
| Answer: \(x^3 + 9x^2 + 20x = 0\) | (B1) |
| Answer: \(x(x^2 + 9x + 20) = 0\) | (M1) |
| Answer: \(x(x+4)(x+5) = 0\) | (A1) |
| Answer: \(\therefore x = 0, x = -4, x = -5\); \(\alpha = 0, \beta = -4\) and \(\gamma = -5\) | (A1) |
| Answer | Marks |
|---|---|
| Answer: If \(\alpha = 0, \beta = -4\) and \(\gamma = -5\), then \(3\alpha = 0, 3\beta = -12\) and \(3\gamma = -15\) | B1 |
| Answer: Therefore \(x(x+12)(x+15) = 0\). Expanding, \((x^2 + 12x)(x + 15) = 0\) or \((x^2 + 15x)(x + 12) = 0\) or \(x(x^2 + 27x + 180) = 0\) | M1 A1 |
| Answer: \(\therefore x^3 + 27x^2 + 180x = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(3\alpha \cdot 3\beta \cdot 3\gamma = 27\alpha\beta\gamma = 0\) (= \(-\frac{d}{a}\)) | (B1) | |
| Answer: \(3\alpha + 3\beta + 3\gamma = 3(\alpha + \beta + \gamma) = -27\) (= \(-\frac{b}{a}, \frac{b}{a} = 27\)) | (M1) | M1 for attempting either method |
| Answer: \(9\alpha\beta + 9\beta\gamma + 9\gamma\alpha = 9(\alpha\beta + \beta\gamma + \gamma\alpha) = 180\) (= \(\frac{c}{a}\)) | (A1) | A1 for both correct |
| Answer: \(\therefore x^3 + 27x^2 + 180x = 0\) | (A1) | For use of their values above |
**Part a)**
**Method 1**
Answer: Because $\alpha\beta\gamma = 0$, put $\alpha = 0$. Then, $\beta + \gamma = -9$ and $\beta\gamma = 20$. Attempt to solve, $\beta = -4$ and $\gamma = -5$ | B1 B1 M1 A1 | Accept solutions where variables are interchanged throughout
**Method 2**
Answer: Because $\alpha\beta\gamma = 0$, either $\alpha = 0$ or $\beta = 0$ or $\gamma = 0$ | (B1) |
Answer: $ax^3 + bx^2 + cx + d = 0$ becomes $x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = 0$ | (B1) |
Answer: $x^3 + 9x^2 + 20x = 0$ | (B1) |
Answer: $x(x^2 + 9x + 20) = 0$ | (M1) |
Answer: $x(x+4)(x+5) = 0$ | (A1) |
Answer: $\therefore x = 0, x = -4, x = -5$; $\alpha = 0, \beta = -4$ and $\gamma = -5$ | (A1) |
**Part b)**
**Method 1**
Answer: If $\alpha = 0, \beta = -4$ and $\gamma = -5$, then $3\alpha = 0, 3\beta = -12$ and $3\gamma = -15$ | B1 |
Answer: Therefore $x(x+12)(x+15) = 0$. Expanding, $(x^2 + 12x)(x + 15) = 0$ or $(x^2 + 15x)(x + 12) = 0$ or $x(x^2 + 27x + 180) = 0$ | M1 A1 |
Answer: $\therefore x^3 + 27x^2 + 180x = 0$ | A1 |
**Method 2**
Answer: $3\alpha \cdot 3\beta \cdot 3\gamma = 27\alpha\beta\gamma = 0$ (= $-\frac{d}{a}$) | (B1) |
Answer: $3\alpha + 3\beta + 3\gamma = 3(\alpha + \beta + \gamma) = -27$ (= $-\frac{b}{a}, \frac{b}{a} = 27$) | (M1) | M1 for attempting either method
Answer: $9\alpha\beta + 9\beta\gamma + 9\gamma\alpha = 9(\alpha\beta + \beta\gamma + \gamma\alpha) = 180$ (= $\frac{c}{a}$) | (A1) | A1 for both correct
Answer: $\therefore x^3 + 27x^2 + 180x = 0$ | (A1) | For use of their values aboveA cubic equation has roots $\alpha$, $\beta$, $\gamma$ such that
$$\alpha + \beta + \gamma = -9, \quad \alpha\beta + \beta\gamma + \gamma\alpha = 20, \quad \alpha\beta\gamma = 0.$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of $\alpha$, $\beta$ and $\gamma$. [4]
\item Find the cubic equation with roots $3\alpha$, $3\beta$, $3\gamma$.
Give your answer in the form $ax^3 + bx^2 + cx + d = 0$, where $a$, $b$, $c$, $d$ are constants to be determined. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q3 [8]}}