| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.8 This is a Further Maths question requiring systematic polynomial transformation (substituting z = w - 1 and expanding), then solving a biquadratic equation involving complex roots. While the technique is standard for FM students, it requires careful algebraic manipulation across multiple steps and working with complex numbers, placing it moderately above average difficulty. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | DR |
| Answer | Marks |
|---|---|
| w 4 + 3 w 2 + 2 = 0 | M1 |
| Answer | Marks |
|---|---|
| A1 | Substitute z = w – 1 |
| Answer | Marks |
|---|---|
| Alternative method: | M1 |
| A1 | Substitute w = z + 1 into end result |
| Answer | Marks |
|---|---|
| Alternative method using symmetry of roots | B1 |
| Answer | Marks |
|---|---|
| B1 | For sums from original equation and finding the sum of the new roots |
| Answer | Marks | Guidance |
|---|---|---|
| ( 1 ) ( 1 ) 3 6 9 1 2 6 3 + + = + + = − + = | B1 | For sums from original equation and finding the sum of the new roots |
| Answer | Marks |
|---|---|
| (b) | w 2 = − 1 , − 2 |
| Answer | Marks |
|---|---|
| z = i − 1 , 2 i − 1 | M1 |
| Answer | Marks |
|---|---|
| A1 | Solving quadratic equation in w2 (or using their variable) |
Question 2:
2 | (a) | DR
( w − 1 4 ) + 4 ( w − 1 3 ) + . . . . . . = 0
w 4 − 4 w 3 + 6 w 2 − 4 w + 1 + . . . . . . .
w 4 + 3 w 2 + 2 = 0 | M1
M1
A1 | Substitute z = w – 1
Expanding with at least (w−1)4 seen
Convincingly shown AG. Must include = 0 on last line. Brackets must be
fully expanded in working or evidence of collection of like terms
A0 if variable used is not w.
Alternative method: | M1
A1 | Substitute w = z + 1 into end result
( z + 1 4 ) + 3 ( z 2 ) + 1 + ...... = 0
z 4 + 4 z 3 + .......
z 4 + 4 z 3 + 2 + 9 z 1 + 0 z 6 = 0 = 0
Alternative method using symmetry of roots | B1
B1
B1 | For sums from original equation and finding the sum of the new roots
For showing convincingly the sum of new roots in pairs
For the last two
4 , 9 . 1 0 , 6 = − = = − =
( 1 ) 4 4 4 0 + = + = − + =
( 1 ) ( 1 ) 3 6 9 1 2 6 3 + + = + + = − + = | B1 | For sums from original equation and finding the sum of the new roots
For showing convincingly the sum of new roots in pairs
B1
For the last two
B1
F o r ( 1 ) ( 1 ) ( 1 ) 0 + + + =
a n d ( 1 ) ( 1 ) ( 1 ) ( 1 ) 2 + + + + =
[3]
(b) | w 2 = − 1 , − 2
w = i , 2 i
z = i − 1 , 2 i − 1 | M1
M1
A1 | Solving quadratic equation in w2 (or using their variable)
Square rooting their w2, including ±, as long as their w2 not both non-
negative and real.
cao
Answers with no working is 0
[3]
M1
A1
Substitute w = z + 1 into end result2 In this question you must show detailed reasoning.\\
The equation $z ^ { 4 } + 4 z ^ { 3 } + 9 z ^ { 2 } + 10 z + 6 = 0$ has roots $\alpha , \beta , \gamma$ and $\delta$.
\begin{enumerate}[label=(\alph*)]
\item Show that a quartic equation whose roots are $\alpha + 1 , \beta + 1 , \gamma + 1$ and $\delta + 1$ is $w ^ { 4 } + 3 w ^ { 2 } + 2 = 0$.
\item Hence determine the exact roots of the equation $z ^ { 4 } + 4 z ^ { 3 } + 9 z ^ { 2 } + 10 z + 6 = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q2 [6]}}