A-LevelFurther MathematicsRoots of polynomial equations9231/12 · May/June 2018 · Q48 Marks
It is given that the equation x³ – 21x² + kx – 216 = 0, where k is a constant, has real roots a, ar and ar⁻¹.
- (i)Find the numerical values of the roots.[6]
- (ii)Deduce the value of k.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- M1 (i) Uses product of roots.
- A1 (i) (a)(ar)(ar⁻¹) = a³ = 216 ⇒ a = 6
- M1 (i) Uses sum of roots. a + ar + ar⁻¹ = 21
- M1 (i) Substitutes for a and solves quadratic. 6(1 + r + r⁻¹) = 21 ⇒ 2r² - 5r + 2 = 0
- A1 (i) r = 2 or r = 0.5
- A1 (i) Roots are 6, 12, 3
- M1 (ii) Or finds coefficient of x in (x-3)(x-6)(x-12). Or substitutes root into equation
- A1 (ii) k = αβ + αγ + βγ = 6(12)+6(3)+12(3) = 126
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