| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Moderate -0.3 This is a standard textbook exercise on polynomial root transformations requiring substitution w = x - 2, then expanding and simplifying. While it involves algebraic manipulation across multiple steps, the technique is routine and commonly practiced in Core Pure 1, making it slightly easier than average for A-level. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{w = x + 2 \Rightarrow\}\ x = w - 2\) | B1 | Selects method of connecting \(x\) and \(w\) by writing \(x = w - 2\) |
| \((w-2)^3 - 7(w-2)^2 - 12(w-2) + 6 = 0\) | M1 | Applies substitution of \(x = "w-2"\) into equation for all occurrences of \(x\) |
| \((w^3 - 6w^2 + 12w - 8) - 7(w^2 - 4w + 4) - 12(w-2) + 6\) → \(w^3 - 6w^2 + 12w - 8 - 7w^2 + 28w - 28 - 12w + 24 + 6 = w^3 + \ldots w^2 + \ldots w + \ldots\) | M1 | Depends on prior substitution; gathering terms into cubic form; condone poor expanding |
| \(w^3 - 13w^2 + 28w - 6 = 0\) | A1 | At least two of \(p\), \(q\), \(r\) correct |
| A1 | Correct final equation including "\(= 0\)"; must be equation in \(w\) | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\alpha + \beta + \gamma = 7,\ \alpha\beta + \beta\gamma + \alpha\gamma = -12,\ \alpha\beta\gamma = -6\) | B1 | Selects method of giving three correct equations for sum, pair sum, product |
| New sum: \(\alpha+2+\beta+2+\gamma+2 = (\alpha+\beta+\gamma)+6 = 7+6 = 13\) | ||
| New pair sum: \((\alpha+2)(\beta+2)+(\alpha+2)(\gamma+2)+(\beta+2)(\gamma+2)\) \(= (\alpha\beta+\alpha\gamma+\beta\gamma)+4(\alpha+\beta+\gamma)+12 = -12+4\times7+12 = 28\) | M1 | Applies process of finding new sum, pair sum and product; allow slips provided attempting correct expressions |
| New product: \((\alpha+2)(\beta+2)(\gamma+2) = \alpha\beta\gamma + 2(\alpha\beta+\alpha\gamma+\beta\gamma)+4(\alpha+\beta+\gamma)+8\) \(= -6+2\times(-12)+4\times7+8 = 6\) | ||
| \(p = -"13"\), \(q = 28\), \(r = -"6"\) or \(w^3 - "13"w^2 + "28"w - "6" = 0\) | M1 | Choosing \(p = -(\text{new sum})\), \(q = \text{new pair sum}\), \(r = -(\text{new product})\) or forming \(w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})\) |
| \(w^3 - 13w^2 + 28w - 6 = 0\) | A1 | At least two of \(p\), \(q\), \(r\) correct |
| A1 | Correct final equation including "\(= 0\)"; must be equation in \(w\) |
## Question 1:
**Main Method (Substitution):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\{w = x + 2 \Rightarrow\}\ x = w - 2$ | **B1** | Selects method of connecting $x$ and $w$ by writing $x = w - 2$ |
| $(w-2)^3 - 7(w-2)^2 - 12(w-2) + 6 = 0$ | **M1** | Applies substitution of $x = "w-2"$ into equation for all occurrences of $x$ |
| $(w^3 - 6w^2 + 12w - 8) - 7(w^2 - 4w + 4) - 12(w-2) + 6$ → $w^3 - 6w^2 + 12w - 8 - 7w^2 + 28w - 28 - 12w + 24 + 6 = w^3 + \ldots w^2 + \ldots w + \ldots$ | **M1** | Depends on prior substitution; gathering terms into cubic form; condone poor expanding |
| $w^3 - 13w^2 + 28w - 6 = 0$ | **A1** | At least two of $p$, $q$, $r$ correct |
| | **A1** | Correct final equation including "$= 0$"; must be equation in $w$ |
| | **(5)** | |
---
**Alternative Method (Sum, Pair Sum and Product of Roots):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\alpha + \beta + \gamma = 7,\ \alpha\beta + \beta\gamma + \alpha\gamma = -12,\ \alpha\beta\gamma = -6$ | **B1** | Selects method of giving three correct equations for sum, pair sum, product |
| New sum: $\alpha+2+\beta+2+\gamma+2 = (\alpha+\beta+\gamma)+6 = 7+6 = 13$ | | |
| New pair sum: $(\alpha+2)(\beta+2)+(\alpha+2)(\gamma+2)+(\beta+2)(\gamma+2)$ $= (\alpha\beta+\alpha\gamma+\beta\gamma)+4(\alpha+\beta+\gamma)+12 = -12+4\times7+12 = 28$ | **M1** | Applies process of finding new sum, pair sum and product; allow slips provided attempting correct expressions |
| New product: $(\alpha+2)(\beta+2)(\gamma+2) = \alpha\beta\gamma + 2(\alpha\beta+\alpha\gamma+\beta\gamma)+4(\alpha+\beta+\gamma)+8$ $= -6+2\times(-12)+4\times7+8 = 6$ | | |
| $p = -"13"$, $q = 28$, $r = -"6"$ **or** $w^3 - "13"w^2 + "28"w - "6" = 0$ | **M1** | Choosing $p = -(\text{new sum})$, $q = \text{new pair sum}$, $r = -(\text{new product})$ or forming $w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})$ |
| $w^3 - 13w^2 + 28w - 6 = 0$ | **A1** | At least two of $p$, $q$, $r$ correct |
| | **A1** | Correct final equation including "$= 0$"; must be equation in $w$ |
---
**Key Notes:**
- Allow a variable other than $w$ for the first 4 marks
- The "$= 0$" is not required until the final mark
- If $x = w - 2$ stated but $w + 2$ substituted, possible to score **B1 M0 M1**
- If $x = w + 2$ stated but $w - 2$ substituted, allow recovery
- In all methods, the final A mark depends on all previous marks\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$x ^ { 3 } - 7 x ^ { 2 } - 12 x + 6 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, determine a cubic equation whose roots are ( $\alpha + 2$ ), $( \beta + 2 )$ and $( \gamma + 2 )$, giving your answer in the form $w ^ { 3 } + p w ^ { 2 } + q w + r = 0$, where $p , q$ and $r$ are integers to be found.
\hfill \mbox{\textit{Edexcel CP1 2023 Q1 [5]}}