A-LevelFurther MathematicsRoots of polynomial equations9231/12 · Oct/Nov 2023 · Q410 Marks

The cubic equation 27x^3 +18x^2 + 6x-1 = 0 has roots α, β, γ.

(a)Show that a cubic equation with roots 3α+1, 3β+1, 3γ+1 is y^3 - y^2 +y-2 = 0.[3]
(b)The sum (3α + 1)^n + (3β+1)^n + (3γ + 1)^n is denoted by S_n. Find the values of S_2 and S_3.[4]
(c)Find the values of S_{-1} and S_{-2}.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[B1] Substitutes x = (y-1)/3. 27((y-1)/3)³ + 18((y-1)/3)² + 6((y-1)/3) - 1 = 0 \implies (y-1)³ + 2(y-1)² + 2(y-1) - 1 = 0.
[M1] Expands. (y³ - 3y² + 3y - 1) + 2(y² - 2y + 1) + 2(y - 1) - 1 = 0
[A1] AG. y³ - y² + y - 2 = 0.
[M1 A1] Uses formula for sum of squares, AG. S₂ = (\sum p)² - 2\sum pq = 1² - 2(1) = -1.
[M1 A1] Uses y³ = y² – y + 2 or expands and uses original equation. S₃ - S₂ + S₁ - 2S₀ = 0 \implies S₃ - (-1) + 1 - 2(3) = 0 \implies S₃ = 4.
[B1] S₋₁ = (Σ pq)/pqr = 1/2.
[M1 A1] Uses 2y^{-2} = y -1+ y^{-1}. 2S_{-2} = S₁ - S₀ + S_{-1} = 1 - 3 + ½ = -3/2. S_{-2} = -3/4. CAO.

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Topic: Roots of polynomial equations, A-Level Further Mathematics (syllabus 9231). Structured question, 10 marks.
Source: A-Level Further Mathematics, Oct/Nov 2023 Paper 1 Variant 2, question 4.

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