AS October 2020 Paper 1 Q9

EdexcelCurrent spec6 marksUsing Roots of Polynomials

9. The cubic equation

\[3x^3 + x^2 - 4x + 1 = 0\]

has roots \(\alpha\), \(\beta\), and \(\gamma\).

Without solving the cubic equation,

(a) determine the value of \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\) (3)

(b) find a cubic equation that has roots \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\) and \(\dfrac{1}{\gamma}\), giving your answer in the form \(x^3 + ax^2 + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers to be determined. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\alpha\beta\gamma = -\dfrac{1}{3}\) and \(\alpha\beta + \alpha\gamma + \beta\gamma = -\dfrac{4}{3}\)	B1	3.1a
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} = \dfrac{-4/3}{-1/3}\)	M1	1.1b
\(= 4\)	A1	1.1b
	(3)

Notes

B1: Correct values for the product and pair sum of the roots

M1: A complete method to find the sum of \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\). Must substitute in their values of the product and pair sum

A1: correct value 4

Note: If candidate does not divide by 3 so that \(\alpha\beta\gamma = -1\) and \(\alpha\beta + \alpha\gamma + \beta\gamma = -4\) the maximum they can score is B0 M1 A0

Mark scheme (b)

Scheme	Marks	AO
\(\left\{\alpha + \beta + \gamma = -\dfrac{1}{3}\right\}\) New product \(= \dfrac{1}{\alpha} \times \dfrac{1}{\beta} \times \dfrac{1}{\gamma} = \dfrac{1}{\alpha\beta\gamma} = \dfrac{1}{-1/3} = \ldots(-3)\) New pair sum \(\dfrac{1}{\alpha\beta} + \dfrac{1}{\beta\gamma} + \dfrac{1}{\alpha\gamma} = \dfrac{\gamma + \alpha + \beta}{\alpha\beta\gamma} = \dfrac{-1/3}{-1/3} = \ldots(1)\)	M1	3.1a
\(x^3 - (\text{part (a)})x^2 + (\text{new pair sum})x - (\text{new product})\,(= 0)\)	M1	1.1b
\(x^3 - 4x^2 + x + 3 = 0\)	A1	1.1b
	(3)
	(6 marks)

Notes

M1: A correct method to find the value of the new pair sum and the value of the new product

M1: Applies \(x^3 - (\text{part (a)})x^2 + (\text{their new pair sum})x - (\text{their new product})\,(= 0)\)

A1: Fully correct equation, in any variable, including = 0

Alternative

Scheme	Marks	AO
e.g. \(z = \dfrac{1}{x} \Rightarrow \dfrac{3}{x^3} + \dfrac{1}{x^2} - \dfrac{4}{x} + 1 = 0\)	M1	3.1a
\(x^3 - 4x^2 + x + 3 = 0\)	M1 A1	1.1b 1.1b
	(3)

M1: Realises the connection between the roots and substitutes into the cubic equation

M1: Manipulates their equation into the form \(x^3 + ax^2 + bx + c = 0\)

A1: Fully correct equation in any variable, including = 0