Factorise polynomials with real coefficients over the complex field, apply the conjugate root theorem and the fundamental theorem of algebra, and find all roots of polynomial equations including those with complex coefficients

What this dot point is asking

QCAA wants you to find every root of a polynomial, real or complex, and to factorise it completely over the complex field. The key tools are the fundamental theorem of algebra, which guarantees the right number of roots, and the conjugate root theorem, which says complex roots of a real polynomial come in conjugate pairs. This lets you use one known complex root to unlock the rest.

The answer

Fundamental theorem of algebra

Every polynomial of degree $n \geq 1$ with complex coefficients has exactly $n$ roots in the complex numbers, counted with multiplicity. Consequently a degree-$n$ polynomial factorises completely into $n$ linear factors over $\mathbb{C}$:

Conjugate root theorem

If a polynomial has real coefficients and $z = \alpha + i\beta$ is a root, then its conjugate $\bar z = \alpha - i\beta$ is also a root. Complex roots of real polynomials therefore occur in conjugate pairs. This is why a real cubic has either three real roots or one real root plus a conjugate pair; complex roots cannot appear alone.

Real quadratic factors from a conjugate pair

A conjugate pair $\alpha \pm i\beta$ multiplies to a quadratic with real coefficients:

So every real polynomial factorises over the reals into real linear factors and real irreducible quadratics, and each such quadratic splits into a conjugate pair over $\mathbb{C}$.

Strategy for finding all roots

Given one complex root of a real polynomial, immediately write down its conjugate as a second root. Multiply the two corresponding factors to get a real quadratic, divide the polynomial by that quadratic (polynomial long division), and solve the remaining lower-degree quotient. This reduces a quartic to a quadratic in one division.

Sum and product of roots

For $P(z) = a_n z^n + \dots + a_0$, the sum of the roots is $-a_{n-1}/a_n$ and the product is $(-1)^n a_0/a_n$. These relations check your factorisation: the roots you find must reproduce the coefficients.

Complex coefficients

When coefficients are themselves complex, the conjugate root theorem no longer applies, since it requires real coefficients. You then find roots directly, for example by the quadratic formula (which holds over $\mathbb{C}$) or by substitution, without assuming conjugate pairs.

The factor and remainder theorems

The factor theorem extends unchanged to the complex field: $z = c$ is a root of $P(z)$ exactly when $(z - c)$ is a factor, and the remainder on dividing $P(z)$ by $(z - c)$ is $P(c)$. This is the engine of every full-factorisation question: once a root is confirmed by substitution, the corresponding linear factor can be peeled off and the degree drops by one. With a conjugate pair you peel off a real quadratic and the degree drops by two, which is why a quartic with one known complex root reduces to a single quadratic after one division.

Multiplicity and the count of roots

The fundamental theorem of algebra counts roots with multiplicity, so a repeated factor must be counted as often as it appears. The polynomial $(z - 1)^2(z^2 + 4)$ has degree four and exactly four roots: $z = 1$ twice and $z = \pm 2i$. If a question states that a real polynomial has a repeated root, that root satisfies both $P(z) = 0$ and $P'(z) = 0$, which gives a second equation to pin it down. Keeping careful track of multiplicity matters when a question asks for the full factorisation rather than just the distinct roots, because dropping a repeated factor leaves the degree wrong and the coefficient check failing.