| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2018 |
| 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 w = 2z + 1, then z = (w-1)/2 into the original equation, followed by algebraic manipulation to clear fractions. It's a routine A-level Further Maths technique with straightforward execution, making it slightly easier than average overall but typical for Core Pure AS. |
| Spec | 4.05b Transform equations: substitution for new roots |
| VILU SIHI NI IIIUM ION OC | VGHV SIHILNI IMAM ION OO | VJYV SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(w = 2z+1 \Rightarrow z = \frac{w-1}{2}\) | B1 | Selects method of connecting \(z\) and \(w\) |
| \(\left(\frac{w-1}{2}\right)^3 - 3\left(\frac{w-1}{2}\right)^2 + \left(\frac{w-1}{2}\right) + 5 = 0\) | M1 | Substitutes \(z = \frac{w-1}{2}\) into \(z^3 - 3z^2 + z + 5 = 0\); allow \(z = 2w+1\) |
| \(\frac{1}{8}(w^3-3w^2+3w-1) - \frac{3}{4}(w^2-2w+1) + \frac{w-1}{2} + 5 = 0\) | M1 | Manipulates into form \(w^3 + pw^2 + qw + r = 0\) |
| \(w^3 - 9w^2 + 19w + 29 = 0\) | A1, A1 | At least two of \(p,q,r\) correct; fully correct equation including "\(= 0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha+\beta+\gamma = 3,\ \alpha\beta+\beta\gamma+\alpha\gamma = 1,\ \alpha\beta\gamma = -5\) | B1 | Selects method of giving three correct equations |
| New sum \(= 2(\alpha+\beta+\gamma)+3 = 9\) | M1 | Applies process of finding new sum, new pair sum, new product |
| New pair sum \(= 4(\alpha\beta+\beta\gamma+\gamma\alpha)+4(\alpha+\beta+\gamma)+3 = 19\) | M1 | Applies \(w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})(=0)\) |
| New product \(= 8\alpha\beta\gamma + 4(\alpha\beta+\beta\gamma+\gamma\alpha)+2(\alpha+\beta+\gamma)+1 = -29\) | ||
| \(w^3 - 9w^2 + 19w + 29 = 0\) | A1, A1 | At least two of \(p,q,r\) correct; fully correct equation including "\(= 0\)" |
# Question 2:
## Main Method
| Answer | Marks | Guidance |
|--------|-------|----------|
| $w = 2z+1 \Rightarrow z = \frac{w-1}{2}$ | B1 | Selects method of connecting $z$ and $w$ |
| $\left(\frac{w-1}{2}\right)^3 - 3\left(\frac{w-1}{2}\right)^2 + \left(\frac{w-1}{2}\right) + 5 = 0$ | M1 | Substitutes $z = \frac{w-1}{2}$ into $z^3 - 3z^2 + z + 5 = 0$; allow $z = 2w+1$ |
| $\frac{1}{8}(w^3-3w^2+3w-1) - \frac{3}{4}(w^2-2w+1) + \frac{w-1}{2} + 5 = 0$ | M1 | Manipulates into form $w^3 + pw^2 + qw + r = 0$ |
| $w^3 - 9w^2 + 19w + 29 = 0$ | A1, A1 | At least two of $p,q,r$ correct; fully correct equation including "$= 0$" |
## ALT 1
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha+\beta+\gamma = 3,\ \alpha\beta+\beta\gamma+\alpha\gamma = 1,\ \alpha\beta\gamma = -5$ | B1 | Selects method of giving three correct equations |
| New sum $= 2(\alpha+\beta+\gamma)+3 = 9$ | M1 | Applies process of finding new sum, new pair sum, new product |
| New pair sum $= 4(\alpha\beta+\beta\gamma+\gamma\alpha)+4(\alpha+\beta+\gamma)+3 = 19$ | M1 | Applies $w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})(=0)$ |
| New product $= 8\alpha\beta\gamma + 4(\alpha\beta+\beta\gamma+\gamma\alpha)+2(\alpha+\beta+\gamma)+1 = -29$ | | |
| $w^3 - 9w^2 + 19w + 29 = 0$ | A1, A1 | At least two of $p,q,r$ correct; fully correct equation including "$= 0$" |\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$z ^ { 3 } - 3 z ^ { 2 } + z + 5 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, find the cubic equation whose roots are ( $2 \alpha + 1$ ), ( $2 \beta + 1$ ) and ( $2 \gamma + 1$ ), 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.
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VILU SIHI NI IIIUM ION OC & VGHV SIHILNI IMAM ION OO & VJYV SIHI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel CP AS 2018 Q2 [5]}}