A2 June 2019 Paper 2 Q2

EdexcelCurrent spec8 marksUsing Roots of Polynomials

2. The roots of the equation

\[x^3 - 2x^2 + 4x - 5 = 0\]

are \(p\), \(q\) and \(r\).

Without solving the equation, find the value of

(i) \(\dfrac{2}{p} + \dfrac{2}{q} + \dfrac{2}{r}\)

(ii) \((p - 4)(q - 4)(r - 4)\)

(iii) \(p^3 + q^3 + r^3\) (8)

Mark scheme (i)

Scheme	Marks	AO
\(p + q + r = 2,\quad pq + pr + qr = 4,\quad pqr = 5\)	B1	3.1a
\(\dfrac{2}{p} + \dfrac{2}{q} + \dfrac{2}{r} = \dfrac{2(pq + pr + qr)}{pqr}\)	M1	1.1b
\(= \dfrac{8}{5}\)	A1ft	1.1b
	(3)

Notes

B1: Identifies the correct values for all 3 expressions (can score anywhere). Allow notation such as \(\sum p,\ \sum pq\) for the sum and pair sum.

M1: Uses a correct identity for the sum

A1ft: Correct value (follow through their 2, 4 and 5)

Alternative for part (i)

Scheme	Marks	AO
\(x = \dfrac{2}{y} \Rightarrow \dfrac{8}{y^3} - \dfrac{8}{y^2} + \dfrac{8}{y} - 5 = 0 \Rightarrow 5y^3 - 8y^2 + 8y - 8 = 0\)	B1	3.1a
\(\dfrac{2}{p} + \dfrac{2}{q} + \dfrac{2}{r} = -\dfrac{-8}{5}\)	M1	1.1b
\(= \dfrac{8}{5}\)	A1ft	1.1b
	(3)

B1: Obtains the correct cubic in “\(y\)”

M1: Uses a correct method

A1ft: Correct value (follow through their 2, 4 and 5)

Mark scheme (ii)

Scheme	Marks	AO
\((p - 4)(q - 4)(r - 4) = (pq - 4p - 4q + 16)(r - 4)\) \(= pqr - 4pq - 4pr - 4qr + 16p + 16q + 16r - 64\)	M1 A1	1.1b 1.1b
\(\bigl(= pqr - 4(pq + pr + qr) + 16(p + q + r) - 64\bigr)\)
\(= 5 - 4(4) + 16(2) - 64 = -43\)	A1	1.1b
	(3)

Notes

M1: Attempt to expand – must have an expression that involves the sum, pair sum and product

A1: Correct expansion

A1: Correct value

Alternative for part (ii)

Scheme	Marks	AO
\((x + 4)^3 - 2(x + 4)^2 + 4(x + 4) - 5 = 0\)	M1	1.1b
\(= \ldots 64 + \ldots - 32 + \ldots 16 + \ldots - 5 = 43\)	A1	1.1b
\(\therefore (p - 4)(q - 4)(r - 4) = -43\)	A1	1.1b
	(3)

M1: Substitutes \(x + 4\) for \(x\) in the given cubic

A1: Calculates the correct constant term

A1: Correct value

Mark scheme (iii)

Scheme	Marks	AO
E.g. \(p^3 + q^3 + r^3 =\) \(= (p + q + r)^3 - 3(p + q + r)(pq + pr + qr) + 3pqr\) or \(= (p + q + r)\bigl((p + q + r)^2 - 2(pq + pr + qr) - pq - pr - qr\bigr) + 3pqr\) or \(= 2\bigl((p + q + r)^2 - 2(pq + pr + qr)\bigr) - 4(p + q + r) + 3pqr\) \(\Rightarrow p^3 + q^3 + r^3 = \ldots\)	M1	3.1a
\(= 2^3 - 3(2)(4) + 3(5) = -1\) \(= 2\left(2^2 - 3(4)\right) + 3(5) = -1\) \(= 2\left(2^2 - 2(4)\right) - 4(2) + 3(5) = -1\)	A1	1.1b
	(2)
	(8 marks)

Notes

M1: Establishes a correct identity that is in terms of the sum, pair sum and product and substitutes to reach a numerical expression for \(p^3 + q^3 + r^3\)

A1: Correct value

Alternative for part (iii)

Scheme	Marks	AO
\(p^3 - 2p^2 + 4p - 5 = 0,\ q^3 - 2q^2 + 4q - 5 = 0,\ r^3 - 2r^2 + 4r - 5 = 0\) \(p^3 + q^3 + r^3 - 2\left(p^2 + q^2 + r^2\right) + 4(p + q + r) - 15 = 0\) \(p^3 + q^3 + r^3 = 2\bigl((p + q + r)^2 - 2(pq + pr + qr)\bigr) - 4(p + q + r) + 15\) \(\Rightarrow p^3 + q^3 + r^3 = \ldots\)	M1	3.1a
\(= 2\left(2^2 - 2(4)\right) - 4(2) + 15 = -1\)	A1	1.1b
	(2)