| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.2 Standard substitution method: let w = x-1, so x = w+1, then substitute into the cubic and expand. This is a routine Further Maths technique with straightforward algebra, typical of specimen/introductory questions. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{w=x-1 \Rightarrow\}\ x=w+1\) | B1 | Selects method of making a connection between \(x\) and \(w\) by writing \(x=w+1\) |
| \((w+1)^3+3(w+1)^2-8(w+1)+6=0\) | M1 | Applies process of substituting \(x=w+1\) into \(x^3+3x^2-8x+6=0\) |
| \(w^3+3w^2+3w+1+3(w^2+2w+1)-8w-8+6=0\) | M1 | Depends on previous M mark. Manipulating equation into form \(w^3+pw^2+qw+r=0\) |
| \(w^3+6w^2+w+2=0\) (at least two of \(p,q,r\) correct) | A1 | |
| \(w^3+6w^2+w+2=0\) (correct final equation) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha+\beta+\gamma=-3,\ \alpha\beta+\beta\gamma+\alpha\gamma=-8,\ \alpha\beta\gamma=-6\) | B1 | Selects method of giving three correct equations each containing \(\alpha\), \(\beta\) and \(\gamma\) |
| Sum roots \(= \alpha-1+\beta-1+\gamma-1 = \alpha+\beta+\gamma-3=-3-3=-6\); pair sum \(= \alpha\beta+\alpha\gamma+\beta\gamma-2(\alpha+\beta+\gamma)+3=-8-2(-3)+3=1\); product \(= \alpha\beta\gamma-(\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha+\beta+\gamma)-1=-6-(-8)-3-1=-2\) | M1 | Applies process of finding sum of roots, pair sum and product |
| Depends on previous M mark. Applies \(w^3 - (\text{sum})w^2+(\text{pair sum})w - \alpha\beta\gamma=0\) | M1 | |
| \(w^3+6w^2+w+2=0\) (at least two of \(p,q,r\) correct) | A1 | |
| \(w^3+6w^2+w+2=0\) (correct final equation) | A1 |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{w=x-1 \Rightarrow\}\ x=w+1$ | B1 | Selects method of making a connection between $x$ and $w$ by writing $x=w+1$ |
| $(w+1)^3+3(w+1)^2-8(w+1)+6=0$ | M1 | Applies process of substituting $x=w+1$ into $x^3+3x^2-8x+6=0$ |
| $w^3+3w^2+3w+1+3(w^2+2w+1)-8w-8+6=0$ | M1 | Depends on previous M mark. Manipulating equation into form $w^3+pw^2+qw+r=0$ |
| $w^3+6w^2+w+2=0$ (at least two of $p,q,r$ correct) | A1 | |
| $w^3+6w^2+w+2=0$ (correct final equation) | A1 | |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha+\beta+\gamma=-3,\ \alpha\beta+\beta\gamma+\alpha\gamma=-8,\ \alpha\beta\gamma=-6$ | B1 | Selects method of giving three correct equations each containing $\alpha$, $\beta$ and $\gamma$ |
| Sum roots $= \alpha-1+\beta-1+\gamma-1 = \alpha+\beta+\gamma-3=-3-3=-6$; pair sum $= \alpha\beta+\alpha\gamma+\beta\gamma-2(\alpha+\beta+\gamma)+3=-8-2(-3)+3=1$; product $= \alpha\beta\gamma-(\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha+\beta+\gamma)-1=-6-(-8)-3-1=-2$ | M1 | Applies process of finding sum of roots, pair sum and product |
| Depends on previous M mark. Applies $w^3 - (\text{sum})w^2+(\text{pair sum})w - \alpha\beta\gamma=0$ | M1 | |
| $w^3+6w^2+w+2=0$ (at least two of $p,q,r$ correct) | A1 | |
| $w^3+6w^2+w+2=0$ (correct final equation) | A1 | |
---\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$x ^ { 3 } + 3 x ^ { 2 } - 8 x + 6 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, find the cubic equation whose roots are $( \alpha - 1 ) , ( \beta - 1 )$ and $( \gamma - 1 )$, 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.\\
(5)
\hfill \mbox{\textit{Edexcel CP AS Q4 [5]}}