Factorise polynomials over the complex numbers using the conjugate root theorem and the fundamental theorem of algebra

What this dot point is asking

SCSA wants you to use these two theorems to find every root of a polynomial and write it as a product of linear factors over $\mathbb{C}$ or real quadratic and linear factors over $\mathbb{R}$.

The fundamental theorem of algebra

This is what makes $\mathbb{C}$ algebraically closed: there is no polynomial equation that fails to have a solution. Over the reals a quadratic with negative discriminant has no roots, but over $\mathbb{C}$ it has two.

The conjugate root theorem

The proof uses that conjugation distributes over sums and products, so if $P(z) = 0$ then $\overline{P(z)} = P(\bar z) = 0$ when the coefficients are real. The immediate consequence: real polynomials of odd degree have at least one real root, and non-real roots always pair up.

Pairing roots into real quadratics

A conjugate pair multiplies to a real quadratic:

So a real polynomial factors over $\mathbb{R}$ into real linear factors (from real roots) and irreducible real quadratics (from conjugate pairs). This is the bridge between the complex factorisation and the real one.

Strategy: from one root to all factors

If you are given or find one non-real root of a real polynomial, you immediately know its conjugate is a root, giving a real quadratic factor. Divide the polynomial by that quadratic to reduce the degree, then factor what remains.

Using the sum and product of roots

For a monic polynomial $z^n + a_{n-1}z^{n-1} + \dots + a_0$, the sum of the roots is $-a_{n-1}$ and the product is $(-1)^n a_0$. These relations give a quick check after factorising and can locate a missing root. For the cubic $z^3 - 5z^2 + 11z - 15$ with roots $3$, $1 + 2i$, $1 - 2i$, the sum is $3 + (1 + 2i) + (1 - 2i) = 5 = -(-5)$ and the product is $3 \times (1^2 + 2^2) = 3 \times 5 = 15 = (-1)^3(-15)$, both confirming the factorisation.

The factor and remainder theorems over the complexes

The factor theorem extends unchanged to the complex numbers: $z = c$ is a root of $P(z)$ exactly when $(z - c)$ is a factor. This is the tool that lets you test a candidate root by substitution and, once confirmed, peel it off as a factor. The remainder theorem likewise says the remainder on dividing $P(z)$ by $(z - c)$ is $P(c)$. These results hold for complex $c$ just as for real $c$, so a complex root found by any means can be turned immediately into a factor, and the degree of the problem drops by one with each factor removed.

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 occurs. The quartic $(z - 2)^2(z^2 + 1)$ has degree $4$ and exactly four roots: $z = 2$ counted twice, and $z = \pm i$. When a question states that a 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 is essential when the question asks for the full factorisation rather than just the distinct roots.

Dividing to reduce the degree

Polynomial long division (or equating coefficients) is the workhorse once a quadratic factor is known. To divide $P(z) = z^3 - 5z^2 + 11z - 15$ by $z^2 - 2z + 5$, write the quotient as $z + c$ and expand $(z^2 - 2z + 5)(z + c) = z^3 + (c - 2)z^2 + (5 - 2c)z + 5c$. Matching the $z^2$ coefficient gives $c - 2 = -5$, so $c = -3$, and the constant $5c = -15$ checks. The remaining linear factor $z - 3$ supplies the last real root.