OCR Further Pure Core AS 2020 November — Question 5 7 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with linearly transformed roots
DifficultyChallenging +1.8 This is a Further Maths question requiring systematic application of Vieta's formulas and algebraic manipulation to transform roots. While the technique is standard for FM students (finding symmetric functions of the new roots in terms of the old), it requires careful multi-step work with sums and products, making it significantly harder than typical A-level questions but not requiring deep novel insight.
Spec4.05a Roots and coefficients: symmetric functions

In this question you must show detailed reasoning. The cubic equation \(5x^3 + 3x^2 - 4x + 7 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\). Find a cubic equation with integer coefficients whose roots are \(\alpha + \beta\), \(\beta + \gamma\) and \(\gamma + \alpha\). [7]

Question 5:
AnswerMarks
53 4 7
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
6
Α + Β + Γ = 2(α + β + γ) = −
5
ΑΒ + ΒΓ + ΓΑ = (α + β)(β + γ) + (β + γ)(γ + α) +
(γ + α)(α + β) = 3(αβ + βγ + γα) + α2 + β2 + γ2
11
= αβ + βγ + γα + (α + β + γ)2 = −
25
ΑΒΓ = (α + β)(β + γ)(γ + α) = 2αβγ + α(αβ + βγ
+ γα) – αβγ + β(αβ + βγ + γα) – αβγ + γ (αβ + βγ
+ γα) – αβγ = (α + β + γ)(αβ + βγ + γα) – αβγ
3 4 7 12 35 47
=− ×− −− = + =
5 5 5 25 25 25
AnswerMarks
a = 25 => 25x3 + 30x2 – 11x – 47 = 0B1
B1ft
M1
A1
M1
A1
AnswerMarks
A11.1a
2.1
2.1
1.1
2.1
1.1
AnswerMarks
1.1For at least 2 correct
2× their Σα
Attempting to find the new Σαβ and
expanding to a symmetrical form
Opening brackets convincingly and
writing in symmetrical form
AnswerMarks
Or any non-zero integer multipleΑ, Β and Γ are the roots of the new
equation
Other correct forms are possible eg
2 2 2 2 2
𝛼𝛼 𝛽𝛽+𝛼𝛼 𝛾𝛾+𝛽𝛽 𝛼𝛼+𝛽𝛽 𝛾𝛾+𝛾𝛾 𝛼𝛼
2
+𝛾𝛾 𝛽𝛽+2𝛼𝛼𝛽𝛽𝛾𝛾
Must be integer coefficients
Alternative method
3
Substitution x=− −u
AnswerMarks
5B1*
B1dep*
AnswerMarks
B1*Used or stated
3  4 7
α+β+γ=− ,  αβ+βγ+γα=− , αβγ=− 
5  5 5
AnswerMarks Guidance
B1dep*Might occur before the first B1 in Note only the sum of roots needed
candidates workinghere
B1**Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
When x=α,
3
u = − −α
5
= α+β+γ−α
AnswerMarks
= α+βShow that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
B1
dep**
AnswerMarks
M13
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
 3 3  3 2
expand − −u and − −u
AnswerMarks
 5   5  3 3  9 27 27 
− −u =−u3+ u2+ u+ 
 5   5 25 125
When x=β/γ,
3
u = − −β/γ
5
= α+β+γ−β/γ
= α+γ/β
 3 3  3 2  3 
5− −u +3− −u −4− −u+7=0
 5   5   5 
( u3+3u2⋅3+3u⋅(3)2+(3)3)
−5
5 5 5
( u2+2u⋅3+(3)2)
+3 +4 ( u+3)+7=0
5 5 5
11 47
−5u3−6u2+ u+ [=0]
5 5
AnswerMarks
25u3+30u2−11u−47=011 47
−5u3−6u2+ u+ [=0]
AnswerMarks
5 5A1
A1Or any non-zero integer multiple
25u3+30u2−11u−47=0
Must be integer coefficients. Must
have “=0”
[7]
B1**
B1
dep**
M1
3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
 3 3  3 2
expand − −u and − −u
 5   5 
 3 3  9 27 27 
− −u =−u3+ u2+ u+ 
 5   5 25 125
A1
A1
Or any non-zero integer multiple
Question 5:
5 | 3 4 7
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
6
Α + Β + Γ = 2(α + β + γ) = −
5
ΑΒ + ΒΓ + ΓΑ = (α + β)(β + γ) + (β + γ)(γ + α) +
(γ + α)(α + β) = 3(αβ + βγ + γα) + α2 + β2 + γ2
11
= αβ + βγ + γα + (α + β + γ)2 = −
25
ΑΒΓ = (α + β)(β + γ)(γ + α) = 2αβγ + α(αβ + βγ
+ γα) – αβγ + β(αβ + βγ + γα) – αβγ + γ (αβ + βγ
+ γα) – αβγ = (α + β + γ)(αβ + βγ + γα) – αβγ
3 4 7 12 35 47
=− ×− −− = + =
5 5 5 25 25 25
a = 25 => 25x3 + 30x2 – 11x – 47 = 0 | B1
B1ft
M1
A1
M1
A1
A1 | 1.1a
2.1
2.1
1.1
2.1
1.1
1.1 | For at least 2 correct
2× their Σα
Attempting to find the new Σαβ and
expanding to a symmetrical form
Opening brackets convincingly and
writing in symmetrical form
Or any non-zero integer multiple | Α, Β and Γ are the roots of the new
equation
Other correct forms are possible eg
2 2 2 2 2
𝛼𝛼 𝛽𝛽+𝛼𝛼 𝛾𝛾+𝛽𝛽 𝛼𝛼+𝛽𝛽 𝛾𝛾+𝛾𝛾 𝛼𝛼
2
+𝛾𝛾 𝛽𝛽+2𝛼𝛼𝛽𝛽𝛾𝛾
Must be integer coefficients
Alternative method
3
Substitution x=− −u
5 | B1*
B1dep*
B1* | Used or stated
3  4 7
α+β+γ=− ,  αβ+βγ+γα=− , αβγ=− 
5  5 5
B1dep* | Might occur before the first B1 in | Note only the sum of roots needed
candidates working | here
B1** | Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
When x=α,
3
u = − −α
5
= α+β+γ−α
= α+β | Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
B1
dep**
M1 | 3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
 3 3  3 2
expand − −u and − −u
 5   5  |  3 3  9 27 27 
− −u =−u3+ u2+ u+ 
 5   5 25 125
When x=β/γ,
3
u = − −β/γ
5
= α+β+γ−β/γ
= α+γ/β
 3 3  3 2  3 
5− −u +3− −u −4− −u+7=0
 5   5   5 
( u3+3u2⋅3+3u⋅(3)2+(3)3)
−5
5 5 5
( u2+2u⋅3+(3)2)
+3 +4 ( u+3)+7=0
5 5 5
11 47
−5u3−6u2+ u+ [=0]
5 5
25u3+30u2−11u−47=0 | 11 47
−5u3−6u2+ u+ [=0]
5 5 | A1
A1 | Or any non-zero integer multiple
25u3+30u2−11u−47=0
Must be integer coefficients. Must
have “=0”
[7]
B1**
B1
dep**
M1
3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
 3 3  3 2
expand − −u and − −u
 5   5 
 3 3  9 27 27 
− −u =−u3+ u2+ u+ 
 5   5 25 125
A1
A1
Or any non-zero integer multiple
In this question you must show detailed reasoning.

The cubic equation $5x^3 + 3x^2 - 4x + 7 = 0$ has roots $\alpha$, $\beta$ and $\gamma$.

Find a cubic equation with integer coefficients whose roots are $\alpha + \beta$, $\beta + \gamma$ and $\gamma + \alpha$. [7]

\hfill \mbox{\textit{OCR Further Pure Core AS 2020 Q5 [7]}}