| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2024 |
| 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 straightforward transformed roots question requiring recognition that roots α-1 come from a horizontal translation, then substituting x+1 into the original polynomial. The transformation identification and algebraic manipulation are standard Further Maths techniques with no novel insight required, making it slightly easier than average. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Translation | M1 | Identifies transformation as a translation; do not accept alternatives such as 'move' |
| Translation \(\begin{bmatrix}-1\\0\end{bmatrix}\) | A1 | Accept "1 to the left" instead of \(\begin{bmatrix}-1\\0\end{bmatrix}\); do not accept any extra erroneous description |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(y+1\) for \(x\) OR states one of: \(\alpha+\beta+\gamma=0\), \(\alpha\beta+\beta\gamma+\gamma\alpha=-1\), \(\alpha\beta\gamma=7\) | B1 | Accept any letter including \(x\) |
| Expands and simplifies \((my+c)^3-(my+c)-7=0\) where \(m\) is non-zero; or correctly calculates two of \(\sum(\alpha-1)\), \(\sum(\alpha-1)(\beta-1)\), \((\alpha-1)(\beta-1)(\gamma-1)\) | M1 | Allow one incorrect simplified coefficient; PI by two correct terms in a cubic equation other than \(x^3\) |
| \((y+1)^3-(y+1)-7=0 \Rightarrow y^3+3y^2+3y+1-y-1-7=0 \Rightarrow y^3+3y^2+2y-7=0\) so \(p(x)=x^3+3x^2+2x-7\) | R1 |
## Question 13(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Translation | M1 | Identifies transformation as a translation; do not accept alternatives such as 'move' |
| Translation $\begin{bmatrix}-1\\0\end{bmatrix}$ | A1 | Accept "1 to the left" instead of $\begin{bmatrix}-1\\0\end{bmatrix}$; do not accept any extra erroneous description |
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## Question 13(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $y+1$ for $x$ OR states one of: $\alpha+\beta+\gamma=0$, $\alpha\beta+\beta\gamma+\gamma\alpha=-1$, $\alpha\beta\gamma=7$ | B1 | Accept any letter including $x$ |
| Expands and simplifies $(my+c)^3-(my+c)-7=0$ where $m$ is non-zero; or correctly calculates two of $\sum(\alpha-1)$, $\sum(\alpha-1)(\beta-1)$, $(\alpha-1)(\beta-1)(\gamma-1)$ | M1 | Allow one incorrect simplified coefficient; PI by two correct terms in a cubic equation other than $x^3$ |
| $(y+1)^3-(y+1)-7=0 \Rightarrow y^3+3y^2+3y+1-y-1-7=0 \Rightarrow y^3+3y^2+2y-7=0$ so $p(x)=x^3+3x^2+2x-7$ | R1 | |
---13 The cubic equation $x ^ { 3 } - x - 7 = 0$ has roots $\alpha , \beta$ and $\gamma$
The cubic equation $\mathrm { p } ( x ) = 0$ has roots $\alpha - 1 , \beta - 1$ and $\gamma - 1$\\
The coefficient of $x ^ { 3 }$ in $\mathrm { p } ( x )$ is 1
13
\begin{enumerate}[label=(\alph*)]
\item Describe fully the transformation that maps the graph of $y = x ^ { 3 } - x - 7$ onto the graph of $y = \mathrm { p } ( x )$\\
13
\item Find $\mathrm { p } ( x )$\\
Turn over for the next question
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q13 [5]}}