Linear factorisation over the complex numbers | Year 11

01 / LEARN

Understand the idea.

Roots and factors

The factor theorem links roots to linear factors: if $p (α) = 0$ , then $z - α$ divides $p (z)$ . A quadratic with roots $α, β$ and leading coefficient $a$ factors as $a (z - α) (z - β)$ .

Keep each root bracketed when forming its factor. For root $- 2 + 3 i$ , subtracting the entire root gives $z - (- 2 + 3 i) = z + 2 - 3 i$ . Substituting the root into its factor should make that factor zero.

Real quadratics can split over the complex numbers

A real quadratic with a negative discriminant has no real linear factorisation, but has conjugate complex linear factors. Completing the square exposes them: $z^{2} + 9 = z^{2} - {(3 i)}^{2} = (z - 3 i) (z + 3 i)$ .

Always retain the original leading coefficient. The roots determine a quadratic only up to a nonzero constant multiplier. Expand the factors to check all coefficients, including the constant term and the coefficient of the highest power.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Factor a shifted quadratic

Factor $z^{2} - 6 z + 10$ into linear factors over the complex numbers.

Worked solution

1
Complete the square.
$z^{2} - 6 z + 10 = {(z - 3)}^{2} + 1$
The quadratic part forms a square after accounting for the constant nine.
2
Find both roots.
$\begin{aligned} {(z - 3)}^{2} & = - 1 \\ z & = 3 \pm i \end{aligned}$
Both signs of i square to negative one.
3
Form one linear factor for each root.
$z^{2} - 6 z + 10 = (z - 3 - i) (z - 3 + i)$
Subtract the entire root from z in each factor; the original polynomial is monic.
4
Expand to verify the identity.
$(z - 3 - i) (z - 3 + i) = {(z - 3)}^{2} - i^{2} = z^{2} - 6 z + 10$
The conjugate product returns the original polynomial, including its constant term.

Answer

$(z - 3 - i) (z - 3 + i)$ .

EXAMPLE 02

Keep a leading coefficient

Factor $2 z^{2} + 4 z + 10$ over the complex numbers.

Worked solution

1
Extract the leading coefficient and complete the square.
$2 z^{2} + 4 z + 10 = 2 [{(z + 1)}^{2} + 4]$
Factoring out two first preserves the correct scale of the quadratic.
2
Find the roots of the bracket.
$\begin{aligned} {(z + 1)}^{2} & = - 4 \\ z & = - 1 \pm 2 i \end{aligned}$
The outer nonzero constant does not change which inputs give zero.
3
Write the full linear factorisation.
$2 z^{2} + 4 z + 10 = 2 (z + 1 - 2 i) (z + 1 + 2 i)$
Each root gives a linear factor, while the original leading coefficient remains outside.
4
Verify by multiplying the factors.
$\begin{aligned} 2 (z + 1 - 2 i) (z + 1 + 2 i) & = 2 [{(z + 1)}^{2} + 4] \\ = 2 z^{2} + 4 z + 10 \end{aligned}$
Conjugate factors cancel their imaginary terms and recover every original coefficient.

Answer

$2 (z + 1 - 2 i) (z + 1 + 2 i)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Factor $z^{2} + 1$ over the complex numbers.

Show answer & reasoning

$(z - i) (z + i)$ .

The roots are $i$ and $- i$ , giving factors $(z - i) (z + i)$ . Their product is $z^{2} - i^{2} = z^{2} + 1$ .

Keep practising

6 worksheet questions · 19 marks

Open worksheet Worked solutions

Read the questions online

Factor $z^{2} + 16$ into complex linear factors.
2 marks
Factor $z^{2} - 2 z + 5$ over the complex numbers.
3 marks
Factor $3 z^{2} + 12$ over the complex numbers.
3 marks
Expand $(z + 2 - 3 i) (z + 2 + 3 i)$ .
3 marks
A quadratic has leading coefficient $- 2$ and roots $2 \pm i$ . Write it in factored and expanded forms.
4 marks
Factor $(z^{2} + 1) (z^{2} - 4)$ into four linear factors and list its roots.
4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Factor $z^{2} + 16$ into complex linear factors.
2 marks
The roots satisfy $z^{2} = - 16$ and are $\pm 4 i$ .
$(z - 4 i) (z + 4 i)$ .
Factor $z^{2} - 2 z + 5$ over the complex numbers.
3 marks
${(z - 1)}^{2} = - 4$ gives roots $1 \pm 2 i$ .
$(z - 1 - 2 i) (z - 1 + 2 i)$ .
Factor $3 z^{2} + 12$ over the complex numbers.
3 marks
First take out $3$ ; $z^{2} + 4$ has roots $\pm 2 i$ .
$3 (z - 2 i) (z + 2 i)$ .
Expand $(z + 2 - 3 i) (z + 2 + 3 i)$ .
3 marks
Use difference of squares: ${(z + 2)}^{2} - {(3 i)}^{2} = {(z + 2)}^{2} + 9$ .
$z^{2} + 4 z + 13$ .
A quadratic has leading coefficient $- 2$ and roots $2 \pm i$ . Write it in factored and expanded forms.
4 marks
The product of conjugate factors is ${(z - 2)}^{2} + 1$ .
Multiply all coefficients by $- 2$ .
$- 2 (z - 2 - i) (z - 2 + i) = - 2 z^{2} + 8 z - 10$ .
Factor $(z^{2} + 1) (z^{2} - 4)$ into four linear factors and list its roots.
4 marks
The first quadratic has complex roots $\pm i$ .
The second has real roots $\pm 2$ .
Each root supplies one factor.
$(z - i) (z + i) (z - 2) (z + 2)$ ; roots $i, - i, 2, - 2$ .

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, real and complex numbers: Cartesian arithmetic, conjugates, quadratic roots and Argand geometry. Polar form and roots of unity are developed in Year 12.

Official curriculum source ↗

In this lesson

Understand the idea.

Roots and factors

Real quadratics can split over the complex numbers

See the method in action.

Factor a shifted quadratic

Complete the square.

Find both roots.

Form one linear factor for each root.

Expand to verify the identity.

Keep a leading coefficient

Extract the leading coefficient and complete the square.

Find the roots of the bracket.

Write the full linear factorisation.

Verify by multiplying the factors.

Try it yourself.

Check your understanding

Keep practising