Solution of polynomial equations over the complex numbers, the fundamental theorem of algebra, the conjugate root theorem for polynomials with real coefficients, and the full factorisation of real polynomials into linear and irreducible quadratic factors

What this dot point is asking

VCAA wants you to solve polynomial equations completely over $\mathbb{C}$, using the fundamental theorem of algebra to know how many roots to expect, the conjugate root theorem to pair complex roots of real polynomials, and the factor theorem to factorise. The goal is a full factorisation into linear factors over $\mathbb{C}$, or into real linear and irreducible quadratic factors over $\mathbb{R}$.

How many roots, and where

The fundamental theorem of algebra guarantees that any polynomial $P(z)$ of degree $n \ge 1$ with complex coefficients has at least one complex root, and by repeated factoring exactly $n$ roots counted with multiplicity. Over $\mathbb{C}$ it therefore splits completely:

Some of the $z_k$ may coincide (repeated roots) or be real; the count of $n$ always holds.

The conjugate root theorem

If $P$ has real coefficients and $a + bi$ (with $b \ne 0$) is a root, then its conjugate $a - bi$ is also a root. The reason is that conjugation distributes over sums and products, and real coefficients are their own conjugates, so $\overline{P(z)} = P(\overline{z})$; if $P(z) = 0$ then $P(\overline{z}) = 0$ too.

A consequence is parity: complex roots come in pairs, so a real polynomial of odd degree must have at least one real root. A real cubic, for example, has either three real roots or one real root and one conjugate pair.

Factorising over the reals

Multiplying a conjugate pair clears the imaginary part:

a quadratic with real coefficients and negative discriminant, hence irreducible over $\mathbb{R}$. Therefore every real polynomial factorises over $\mathbb{R}$ into real linear factors (one per real root) and irreducible real quadratics (one per conjugate pair). To carry this out:

1. Find one root, by inspection, the rational root test, or a given clue.
2. Divide $P(z)$ by the corresponding factor to reduce the degree.
3. Solve the reduced polynomial, using the conjugate pair to write a quadratic factor directly when a complex root is found.

Examples in context

Example 1. A real polynomial with roots $2i$ and $-2i$ has the factor $z^2 + 4$ (here $a = 0$, $b = 2$).

Example 2. $z^4 + 1 = 0$ has four roots, the fourth roots of $-1$, forming two conjugate pairs and factorising over $\mathbb{R}$ into two irreducible quadratics.

Try this

Q1. A real quartic has roots $1 + i$ and $-3$. Name one further root it must have. [1 mark]

• Cue. $1 - i$, the conjugate of $1 + i$.

Q2. Solve $z^2 + 4z + 13 = 0$ over $\mathbb{C}$. [2 marks]

• Cue. $z = -2 \pm 3i$.

Q3. Given $z = 1 - i$ is a root of $z^3 - 3z^2 + 4z - 2 = 0$, find all roots. [3 marks]

• Cue. Conjugate $1 + i$ gives factor $z^2 - 2z + 2$; division leaves $z - 1$, so roots are $1$, $1 \pm i$.