| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| 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 x = w - 3 into the original equation and simplification. It involves routine algebraic manipulation and application of a well-practiced technique from the A-level syllabus, making it slightly easier than average but requiring careful execution. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{w = x+3 \Rightarrow\}\ x = w-3\) | B1 | Selects method of connecting \(x\) and \(w\) by writing \(x = w-3\) |
| \(2(w-3)^3 + 6(w-3)^2 - 3(w-3) + 12\ (=0)\) | M1 | Applies substitution of \(x = aw \pm b\) into \(2x^3 + 6x^2 - 3x + 12\ (=0)\) |
| \(2w^3 - 18w^2 + 54w - 54 + 6(w^2 - 6w + 9) - 3w + 9 + 12\ (=0)\) | M1 | Manipulates resulting equation into form \(pw^3 + qw^2 + rw + s\ (=0)\), \(p \neq 0\) |
| \(2w^3 - 12w^2 + 15w + 21 = 0\) | A1 | At least three of \(p, q, r, s\) correct with integer coefficients |
| So \(p=2,\ q=-12,\ r=15,\ s=21\) | A1 | Correct final equation including "= 0"; accept integer multiples |
# Question 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $\{w = x+3 \Rightarrow\}\ x = w-3$ | B1 | Selects method of connecting $x$ and $w$ by writing $x = w-3$ |
| $2(w-3)^3 + 6(w-3)^2 - 3(w-3) + 12\ (=0)$ | M1 | Applies substitution of $x = aw \pm b$ into $2x^3 + 6x^2 - 3x + 12\ (=0)$ |
| $2w^3 - 18w^2 + 54w - 54 + 6(w^2 - 6w + 9) - 3w + 9 + 12\ (=0)$ | M1 | Manipulates resulting equation into form $pw^3 + qw^2 + rw + s\ (=0)$, $p \neq 0$ |
| $2w^3 - 12w^2 + 15w + 21 = 0$ | A1 | At least three of $p, q, r, s$ correct with integer coefficients |
| So $p=2,\ q=-12,\ r=15,\ s=21$ | A1 | Correct final equation including "= 0"; accept integer multiples |
---\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$2 x ^ { 3 } + 6 x ^ { 2 } - 3 x + 12 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, find the cubic equation whose roots are $( \alpha + 3 ) , ( \beta + 3 )$ and $( \gamma + 3 )$, giving your answer in the form $p w ^ { 3 } + q w ^ { 2 } + r w + s = 0$, where $p , q , r$ and $s$ are integers to be found.
\hfill \mbox{\textit{Edexcel CP AS 2019 Q2 [5]}}