A2 June 2025 Paper 2 Q2

OCR ACurrent spec4 marksComplex Numbers Finding Roots of Polynomials

2 You are given that \(-7 - 5\mathrm{i}\) is one root of the equation \(x^3 + 10x^2 + 18x - 296 = 0\).

(a) Write down another complex root of the equation \(x^3 + 10x^2 + 18x - 296 = 0\). [1]

(b) Using your answer to part (a), express \(x^3 + 10x^2 + 18x - 296\) as a product of a real linear factor and a real quadratic factor. [3]

Scheme	Marks	AO
\(-7 + 5\mathrm{i}\)	B1	1.2
	[1]

Mark scheme (b)

Scheme	Marks	AO
\((-7 + 5\mathrm{i})(-7 - 5\mathrm{i}) = 49 + 25 = 74\) or \((-7 + 5\mathrm{i}) + (-7 - 5\mathrm{i}) = -14\)	M1	1.1
\(\Rightarrow x^2 + 14x + 74\)	A1	2.2a
\(\Rightarrow x - 4\) (hence \((x - 4)(x^2 + 14x + 74)\))	A1	1.1
	[3]

Notes

M1: Finding sum or product of known roots (could be embedded)
Or dividing cubic by one of \((x - (-7 + 5\mathrm{i}))\), \((x - (-7 - 5\mathrm{i}))\) and the quotient by the other
Or by multiplying \((x - (-7 + 5\mathrm{i}))(x - (-7 - 5\mathrm{i})) = \ldots\)
(corrected from the printed mark scheme: the sum is printed as \((-7 + 5\mathrm{i}) + (-7 + 5\mathrm{i}) = -14\); the two roots are \(-7 + 5\mathrm{i}\) and \(-7 - 5\mathrm{i}\))

A1: No need to reassemble if both factors correct and unambiguous
SCB1 for correct answer without evidence of use of (a) (max 1/3).

Alternative method

Scheme	Marks
\((-7 + 5\mathrm{i})(-7 - 5\mathrm{i}) = 49 + 25 = 74\) \(\frac{296}{74} = 4\)	M1
\(\Rightarrow x - 4\)	A1
\(\Rightarrow x^2 + 14x + 74\) (hence \((x - 4)(x^2 + 14x + 74)\))	A1

M1: Finding product of known roots and \(\alpha\beta\gamma = -\frac{d}{a}\) to find the real root.
(corrected from the printed mark scheme, which has \(\alpha\beta\gamma = -\frac{c}{a}\); for \(ax^3 + bx^2 + cx + d = 0\) the product of the roots is \(-\frac{d}{a}\))

A1: No need to reassemble if both factors correct and unambiguous