2019 H2 Maths Paper 1 Q1 — Worked Solution

Solution

4 marks

Approach

$f$ has real coefficients, so non-real roots occur in conjugate pairs. Since $2 + i$ is a root, $2 - i$ must also be a root. Together with the given root $- 3$ , this accounts for all three roots of the cubic, and $f(z)$ factorises completely in terms of $a$ .

Since $2 + i$ , $2 - i$ and $- 3$ are the three roots of $f(z) = 0$ ,

f(z) = a\left( z - (2 + i) \right)\left( z - (2 - i) \right)(z + 3)

Multiplying the conjugate pair first,

\left( z - (2 + i) \right)\left( z - (2 - i) \right)

= \left( (z - 2) - i \right)\left( (z - 2) + i \right)

= (z - 2)^{2} + 1

= z^{2} - 4z + 5

f(z) = a\left( z^{2} - 4z + 5 \right)(z + 3)

= a\left( z^{3} + 3z^{2} - 4z^{2} - 12z + 5z + 15 \right)

= a\left( z^{3} - z^{2} - 7z + 15 \right)

= az^{3} - az^{2} - 7az + 15a

Comparing with $f(z) = az^{3} + bz^{2} + cz + d$ ,

b = - a,\ \ c = - 7a,\ \ d = 15a

To check your answer: substituting

z = - 3

into

az^{3} - az^{2} - 7az + 15a

gives

- 27a - 9a + 21a + 15a = 0

. ✓

$\therefore$ $b = - a$ , $c = - 7a$ , $d = 15a$ .

2019 H2 Maths Paper 1, Question 1

Solution

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