Specialist Maths — Question 18

QCAA 2022, Paper 2 · 5 marks

Q18 · 2022 · Technology-activeComplex unfamiliar5 marks

QUESTION 18 (5 marks)

Consider the polynomials

P(z)=z^3+(i-a)z^2-2biz+3i

and

Q(z)=z-2i

, where

a,b\in\mathbb R

. Given

P(z)/Q(z)

has a remainder of

a-bi

, evaluate the reasonableness that

(z-(a-bi))

is a factor of

P(z)

WORKED SOLUTION

5 marks

ANSWER

It is not reasonable;

a=-12

b=9

and

P(-12-9i)\ne0

Worked solution

By the remainder theorem,

P(2i)=a-bi

. Direct substitution gives

P(2i)=4a+4b-9i.

Equating real and imaginary parts with

a-bi

gives

a=-12

and

b=9

. Hence

P(a-bi)=P(-12-9i)\approx1566-285i\ne0.

Therefore it is not reasonable that

z-(a-bi)

is a factor of

P(z)

correctly determines an expression for

P(2i)

using the remainder theorem

[1 mark]

correctly determines an expression for

P(2i)

using substitution into

P(z)

[1 mark]

forms two simultaneous equations by equating parts

[1 mark]

determines

P(a-bi)

using the values for

a

and

b

[1 mark]

evaluates the reasonableness of the statement using mathematical reasoning

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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