Linear factorisation over the complex numbers | Year 11

01 / LEARN

The idea, step by step

Roots and factors

If $α$ is a root of a polynomial $p (z)$ , then $z - α$ is a factor. A quadratic with leading coefficient $a$ and roots $α, β$ factors as $a (z - α) (z - β)$ .

Do not reverse the signs inconsistently: the factor for root $- 2 + 3 i$ is $z + 2 - 3 i$ . Substitution of the root into its own factor gives zero and is a quick check.

Real quadratics can split over the complex numbers

A real quadratic with negative discriminant cannot be factored into real linear factors, but it does split into conjugate complex factors. For example, $z^{2} + 9 = (z - 3 i) (z + 3 i)$ .

Expanding the proposed factors checks every coefficient. Keep any original leading coefficient outside the factors; roots alone do not determine it. This stage uses Cartesian form, not polar roots-of-unity techniques.

A common trap

Writing $z^{2} + 4 = (z - 2) (z + 2)$ .

Instead: That product is $z^{2} - 4$ . The required factors use roots $\pm 2 i$ : $(z - 2 i) (z + 2 i)$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Factor a shifted quadratic

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

Complete the square: $z^{2} - 6 z + 10 = {(z - 3)}^{2} + 1$ .
The roots are $3 + i$ and $3 - i$ .
Form a factor $z - α$ for each root.
Multiply the conjugate factors to recover ${(z - 3)}^{2} + 1$ .

(z - 3 - i) (z - 3 + i)

Example 2

Keep a leading coefficient

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

Write the polynomial as $2 [{(z + 1)}^{2} + 4]$ .
Its roots are $- 1 \pm 2 i$ .
The factorisation must retain the multiplier $2$ .
The product inside is ${(z + 1)}^{2} + 4 = z^{2} + 2 z + 5$ .

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

Check your understanding

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

Show answer

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

The roots are $i$ and $- i$ , and the product expands to $z^{2} - i^{2} = z^{2} + 1$ .

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF 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 $i, - i, 2, - 2$ .

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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 ↗

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