Standard +0.3 This is a standard Further Maths question on transformed roots requiring knowledge that if α, β, γ are roots of ax³+bx²+cx+d=0, then 1/α, 1/β, 1/γ are roots of dx³+cx²+bx+a=0 (or using substitution y=1/x). While it's a Further Maths topic, the technique is routine and mechanical once learned, requiring only coefficient reversal or a straightforward substitution—no novel insight or complex multi-step reasoning needed.
2 In this question you must show detailed reasoning.
The cubic equation \(2 x ^ { 3 } + 3 x ^ { 2 } - 5 x + 4 = 0\) has roots \(\alpha , \beta\) and \(\gamma\). By making an appropriate substitution, or otherwise, find a cubic equation with integer coefficients whose roots are \(\frac { 1 } { \alpha } , \frac { 1 } { \beta }\) and \(\frac { 1 } { \gamma }\).
Other letters can be used as variable. If \(x\) used allow B1 for sight of \(2\left(\frac{1}{x}\right)^3+3\left(\frac{1}{x}\right)^2-5\left(\frac{1}{x}\right)+4\ (=0)\)
For substituting \(\frac{1}{u}\) into the given equation and attempting to multiply by \(u^3\). If no \(u^3\) outside brackets then need to see at least two terms multiplied by \(u^3\).
\(4u^3-5u^2+3u+2=0\)
A1
Or multiple of this (with integer coefficients). Condone \(x\) as variable. Must have "\(=0\)"
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u=\frac{1}{x}$ | B1 | Other letters can be used as variable. If $x$ used allow B1 for sight of $2\left(\frac{1}{x}\right)^3+3\left(\frac{1}{x}\right)^2-5\left(\frac{1}{x}\right)+4\ (=0)$ |
| $u^3\left(2\left(\frac{1}{u}\right)^3+3\left(\frac{1}{u}\right)^2-5\left(\frac{1}{u}\right)+4\right)=0$ | M1 | For substituting $\frac{1}{u}$ into the given equation and attempting to multiply by $u^3$. If no $u^3$ outside brackets then need to see at least two terms multiplied by $u^3$. |
| $4u^3-5u^2+3u+2=0$ | A1 | Or multiple of this (with integer coefficients). Condone $x$ as variable. Must have "$=0$" |
**Alt (using Vieta's):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha+\beta+\gamma=-\frac{3}{2}$; $\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{5}{2}$; $\alpha\beta\gamma=-2$ | M1 | Allow one sign slip |
| $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac{5}{4}$; $\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}=\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}=\frac{3}{4}$; $\frac{1}{\alpha\beta\gamma}=-\frac{1}{2}$ | M1 | At least 2 correct |
| $4x^3-5x^2+3x+2=0$ | A1 | Must be integer coefficients, must have "$=0$" |
**[3 marks total]**
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2 In this question you must show detailed reasoning.\\
The cubic equation $2 x ^ { 3 } + 3 x ^ { 2 } - 5 x + 4 = 0$ has roots $\alpha , \beta$ and $\gamma$. By making an appropriate substitution, or otherwise, find a cubic equation with integer coefficients whose roots are $\frac { 1 } { \alpha } , \frac { 1 } { \beta }$ and $\frac { 1 } { \gamma }$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q2 [3]}}