| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.3 This is a standard transformed roots question requiring substitution and algebraic manipulation. Students use the relationship w = 3x - 2, so x = (w+2)/3, substitute into the original equation, and simplify. While it involves careful algebra with fractions, it's a routine technique taught explicitly in Core Pure AS with no novel problem-solving required. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(w = 3x - 2 \Rightarrow x = \frac{w+2}{3}\) | B1 | 3.1a – Selects method of connecting \(x\) and \(w\) |
| \(9\left(\frac{w+2}{3}\right)^3 - 5\left(\frac{w+2}{3}\right)^2 + 4\left(\frac{w+2}{3}\right) + 7 = 0\) | M1 | 3.1a – Applies substitution of \(x = \frac{w+2}{3}\) into \(9x^3 - 5x^2 + 4x + 7 = 0\) |
| \(\frac{1}{3}(w^3+6w^2+12w+8) - \frac{5}{9}(w^2+4w+4) + \frac{4}{3}(w+2) + 7 = 0\) | Expansion step | |
| \(3w^3 + 13w^2 + 28w + 91 = 0\) | dM1, A1, A1 | 1.1b – dM1: manipulates to \(aw^3+bw^2+cw+d=0\); A1: at least two of \(a,b,c,d\) correct; A1: fully correct equation in terms of \(w\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\alpha+\beta+\gamma = \frac{5}{9},\ \alpha\beta+\beta\gamma+\alpha\gamma = \frac{4}{9},\ \alpha\beta\gamma = -\frac{7}{9}\) | B1 | 3.1a – Three correct equations |
| New sum \(= 3(\alpha+\beta+\gamma) - 6 = -\frac{13}{3}\) | ||
| New pair sum \(= 9(\alpha\beta+\beta\gamma+\gamma\alpha) - 12(\alpha+\beta+\gamma) + 12 = \frac{28}{3}\) | M1 | 3.1a – Finds new sum, pair sum, product |
| New product \(= 27\alpha\beta\gamma - 18(\alpha\beta+\beta\gamma+\gamma\alpha) + 12(\alpha+\beta+\gamma) - 8 = -\frac{91}{3}\) | ||
| \(w^3 - \left(-\frac{13}{3}\right)w^2 + \frac{28}{3}w - \left(-\frac{91}{3}\right) = 0\) | dM1 | 1.1b |
| \(3w^3 + 13w^2 + 28w + 91 = 0\) | A1, A1 | 1.1b – A1: at least two correct; A1: fully correct |
# Question 2:
## Main Method:
| Working | Mark | Notes |
|---------|------|-------|
| $w = 3x - 2 \Rightarrow x = \frac{w+2}{3}$ | B1 | 3.1a – Selects method of connecting $x$ and $w$ |
| $9\left(\frac{w+2}{3}\right)^3 - 5\left(\frac{w+2}{3}\right)^2 + 4\left(\frac{w+2}{3}\right) + 7 = 0$ | M1 | 3.1a – Applies substitution of $x = \frac{w+2}{3}$ into $9x^3 - 5x^2 + 4x + 7 = 0$ |
| $\frac{1}{3}(w^3+6w^2+12w+8) - \frac{5}{9}(w^2+4w+4) + \frac{4}{3}(w+2) + 7 = 0$ | | Expansion step |
| $3w^3 + 13w^2 + 28w + 91 = 0$ | dM1, A1, A1 | 1.1b – dM1: manipulates to $aw^3+bw^2+cw+d=0$; A1: at least two of $a,b,c,d$ correct; A1: fully correct equation in terms of $w$ |
## Alternative Method (Vieta's):
| Working | Mark | Notes |
|---------|------|-------|
| $\alpha+\beta+\gamma = \frac{5}{9},\ \alpha\beta+\beta\gamma+\alpha\gamma = \frac{4}{9},\ \alpha\beta\gamma = -\frac{7}{9}$ | B1 | 3.1a – Three correct equations |
| New sum $= 3(\alpha+\beta+\gamma) - 6 = -\frac{13}{3}$ | | |
| New pair sum $= 9(\alpha\beta+\beta\gamma+\gamma\alpha) - 12(\alpha+\beta+\gamma) + 12 = \frac{28}{3}$ | M1 | 3.1a – Finds new sum, pair sum, product |
| New product $= 27\alpha\beta\gamma - 18(\alpha\beta+\beta\gamma+\gamma\alpha) + 12(\alpha+\beta+\gamma) - 8 = -\frac{91}{3}$ | | |
| $w^3 - \left(-\frac{13}{3}\right)w^2 + \frac{28}{3}w - \left(-\frac{91}{3}\right) = 0$ | dM1 | 1.1b |
| $3w^3 + 13w^2 + 28w + 91 = 0$ | A1, A1 | 1.1b – A1: at least two correct; A1: fully correct |
**Total: 5 marks**
---\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$9 x ^ { 3 } - 5 x ^ { 2 } + 4 x + 7 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, find the cubic equation whose roots are ( $3 \alpha - 2$ ), ( $3 \beta - 2$ ) and ( $3 \gamma - 2$ ), giving your answer in the form $a w ^ { 3 } + b w ^ { 2 } + c w + d = 0$, where $a , b , c$ and $d$ are integers to be determined.
\hfill \mbox{\textit{Edexcel CP AS 2021 Q2 [5]}}