A2 June 2024 Paper 1 Q2

EdexcelCurrent spec8 marksUsing Roots of Polynomials

2. The roots of the equation

\[2x^3 - 3x^2 + 12x + 7 = 0\]

are \(\alpha\), \(\beta\) and \(\gamma\)

Without solving the equation,

(a) write down the value of each of\[\alpha + \beta + \gamma \qquad \alpha\beta + \alpha\gamma + \beta\gamma \qquad \alpha\beta\gamma\] (1)

(b) Use the answers to part (a) to determine the value of

(i) \(\dfrac{2}{\alpha} + \dfrac{2}{\beta} + \dfrac{2}{\gamma}\)

(ii) \((\alpha - 1)(\beta - 1)(\gamma - 1)\)

(iii) \(\alpha^2 + \beta^2 + \gamma^2\) (7)

Scheme	Marks	AO
\(\alpha + \beta + \gamma = \dfrac{3}{2},\ \alpha\beta + \alpha\gamma + \beta\gamma = 6,\ \alpha\beta\gamma = -\dfrac{7}{2}\)	B1	1.1b
	(1)

Mark scheme (b)(i)

Scheme	Marks	AO
\(\dfrac{2}{\alpha} + \dfrac{2}{\beta} + \dfrac{2}{\gamma} = \dfrac{2(\alpha\beta + \alpha\gamma + \beta\gamma)}{\alpha\beta\gamma} = 2 \times \dfrac{\text{``}6\text{''}}{\text{``}{-7/2}\text{''}}\)	M1	1.1b
\(= -\dfrac{24}{7}\) oe	A1ft	1.1b

Notes

(b) Note question requires use of (a) so other methods will score no marks.

M1: Use a correct identity with an attempt to substitute their values into the correct places (allowing a slip) to find the value required. If identity is not shown it may be implied by the working.

A1ft: Correct value (follow through their part (a)) Note that this means the error. \(\Sigma\alpha_i = 3,\ \Sigma\alpha_i\beta_j = 12,\ \Pi\alpha_i = -7\) will score M1A1ft for the correct answer albeit from incorrect values.

Mark scheme (b)(ii)

Scheme	Marks	AO
\((\alpha - 1)(\beta - 1)(\gamma - 1) = \bigl(\alpha\beta - (\alpha + \beta) + 1\bigr)(\gamma - 1) = \ldots\)	M1	1.1b
\(= \alpha\beta\gamma - (\alpha\beta + \alpha\gamma + \beta\gamma) + \alpha + \beta + \gamma - 1\)	A1	1.1b
\(= -9\)	A1	1.1b

Notes

M1: Attempts to expand the product fully (allow sign slips and at most one incorrect or missing term).

A1: Correct expansion in terms of product, pair sum and sum - must be seen grouped or implied by substitution of values seen.

A1: Correct value.

Mark scheme (b)(iii)

Scheme	Marks	AO
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)\) \(= \left(\text{``}\dfrac{3}{2}\text{''}\right)^2 - 2\text{``}6\text{''}\)	M1	3.1a
\(= \dfrac{9}{4} - 2(6) = -\dfrac{39}{4}\) oe	A1ft	1.1b
	(7)
	(8 marks)

Notes

A1ft: Correct value from correct working (follow through their part (a)). Note this means the error \(\Sigma\alpha_i = -\dfrac{3}{2}\) will still give the A1ft here if correct identity is used.