AS June 2018 Paper 1 Q5

OCR ACurrent spec10 marksComplex Numbers Finding Roots of Polynomials

5 In this question you must show detailed reasoning.

(i) Express \((2 + 3\mathrm{i})^3\) in the form \(a + \mathrm{i}b\). [3]

(ii) Hence verify that \(2 + 3\mathrm{i}\) is a root of the equation \(3z^3 - 8z^2 + 23z + 52 = 0\). [3]

(iii) Express \(3z^3 - 8z^2 + 23z + 52\) as the product of a linear factor and a quadratic factor with real coefficients. [4]

Mark scheme (i)

Scheme	Marks	AO
\(2^3 + 3 \times 2^2 \times 3\mathrm{i} + 3 \times 2 \times (3\mathrm{i})^2 + (3\mathrm{i})^3\)	M1	1.1
\(2^3 + 3 \times 2^2 \times 3\mathrm{i} - 3 \times 2 \times 3^2 - 3^3\mathrm{i}\) or better	A1	1.1
\(-46 + 9\mathrm{i}\)	A1	1.1
	[3]

Notes

M1: Binomial expansion. Must be 4 terms with 1, 3, 3, 1 soi and correct powers. Condone missing brackets
Or by \((2 + 3\mathrm{i})^2 \times (2 + 3\mathrm{i})\) but marks only to awarded once all binomial brackets expanded.

A1: (1st) All correct and \(\mathrm{i}^2 = -1\) twice.
Must see evidence of \(\mathrm{i}^2\) becoming \(-1\) (could be in a table, expanding brackets etc)

A1: (2nd) SC if the only working seen is \((-5 + 12\mathrm{i})(2 + 3\mathrm{i}) = -46 + 9\mathrm{i}\) award B1

Mark scheme (ii)

Scheme	Marks	AO
\((2 + 3\mathrm{i})^2 = -5 + 12\mathrm{i}\)	B1	1.1
\(3(-46 + 9\mathrm{i}) - 8(-5 + 12\mathrm{i}) + 23(2 + 3\mathrm{i}) + 52\)	M1	3.1a
\(= -138 + 27\mathrm{i} + 40 - 96\mathrm{i} + 46 + 69\mathrm{i} + 52\) \(= -138 + 138 + 96\mathrm{i} - 96\mathrm{i} = 0\)	A1 (AG)	1.1
	[3]

Notes

B1: May be seen in (i). May be in working below.
If only seen in (i) and not implied by working for this part award B0

M1: Attempt to substitute their \(z^2\) and \(z^3\) into \(3z^3 - 8z^2 + 23z + 52\)
\(3(2 + 3\mathrm{i})^3 - 8(2 + 3\mathrm{i})^2 + 23(2 + 3\mathrm{i}) + 52\) enough for M1

A1: Convincingly cancels to 0
Must show some collection/cancellation Could be gathering real and imaginary terms

Mark scheme (iii)

Scheme	Marks	AO
\(2 - 3\mathrm{i}\) is also a root	B1	1.1
\((z - (2 + 3\mathrm{i}))(z - (2 - 3\mathrm{i}))\)	M1	2.2a
\(= z^2 - 4z + 13\)	A1	1.1
\((z^2 - 4z + 13)(3z + 4)\)	A1	2.2a
	[4]

Notes

B1: Seen or implied

M1: ...is the required quadratic factor

A1: (2nd) Can be deduced by inspection
Must be written as a product of linear factor and quadratic factor
Condone “= 0” present.

Alternative

Scheme	Marks
\(2 - 3\mathrm{i}\) is also a root	B1
\(z - 2 = \pm 3\mathrm{i}\) \((z - 2)^2 = -9\)	M1
\(z^2 - 4z + 13 (= 0)\)	A1
\((z^2 - 4z + 13)(3z + 4)\)	A1

B1: Seen or implied

M1: Using the two roots to get an equation in \(z^2\)

A1: (2nd) Can be deduced by inspection
Condone “= 0” present.