Solving of polynomials over the complex numbers
Learn how to factorise polynomials over \(\mathbb{C}\) in Year 12 Specialist Mathematics for Queensland (QCAA). Building on complex arithmetic, you use the factor and remainder theorems to find factors and the complex conjugate root theorem to pair non-real zeros.
You will factorise cubics and quartics fully into linear factors, turn a conjugate pair into a real quadratic factor, and solve polynomial equations of order up to four over \(\mathbb{C}\) — underpinned by the fundamental theorem of algebra, which guarantees every zero.
Theory
Factorisation of polynomials over \(\mathbb{C}\) breaks a polynomial into linear factors using the complex numbers, a core skill of Year 12 Specialist Mathematics (QCAA, Queensland, Unit 3). The factor and remainder theorems locate a factor, the complex conjugate root theorem pairs non-real zeros, and the fundamental theorem of algebra guarantees a degree-\(n\) polynomial has exactly \(n\) zeros over \(\mathbb{C}\).
The remainder theorem says the remainder when a polynomial \(P(z)\) is divided by \(z-a\) is simply \(P(a)\). The factor theorem is the special case that matters most: \((z-a)\) is a factor of \(P(z)\) exactly when \(P(a)=0\), that is, when \(a\) is a zero.
The fundamental theorem of algebra guarantees that a polynomial of degree \(n\) has exactly \(n\) zeros over \(\mathbb{C}\) (counting multiplicity), so it factors into \(n\) linear factors over the complex numbers. A real quadratic with no real zeros, such as \(z^2+1\), does factor over \(\mathbb{C}\) as \((z-i)(z+i)\).
The complex conjugate root theorem applies when the coefficients are real: if \(a+bi\) is a zero then its conjugate \(a-bi\) is a zero too. Non-real zeros therefore come in conjugate pairs, and each pair multiplies to a real quadratic factor \(z^2-2az+(a^2+b^2)\).
The theorem needs those real coefficients. If a polynomial has imaginary coefficients its non-real zeros need not pair up, so the conjugate of a zero is not guaranteed to be a zero — you must factor directly instead.
The remainder when \(P(z)\) is divided by \(z-a\), and the factor test:
By the fundamental theorem of algebra, a monic degree-\(n\) polynomial splits into \(n\) linear factors over \(\mathbb{C}\):
A conjugate pair \(a\pm bi\) (real coefficients) multiplies to a real quadratic factor:
How to factorise a polynomial over \(\mathbb{C}\)
- Find one zero. Test small values with the factor theorem (\(P(a)=0\)), spot a grouping, or use a zero the question gives you.
- Pair a non-real zero. If the coefficients are real and \(a+bi\) is a zero, its conjugate \(a-bi\) is too; multiply them to a real quadratic \(z^2-2az+(a^2+b^2)\).
- Divide out. Divide \(P(z)\) by the known factor (linear or quadratic) to reduce the degree, then factorise the quotient.
- Write linear factors and zeros. Split every quadratic over \(\mathbb{C}\) and list all zeros, keeping surds and \(i\) exact.
Test \(z=1\) for the factor, then evaluate \(P(-1)\) for the remainder:
| \(P(1)\) | \(=\) | \((1)^3-2(1)^2-5(1)+6\) |
| \(=\) | \(1-2-5+6\) | |
| \(=\) | \(0\) | |
| \(P(-1)\) | \(=\) | \((-1)^3-2(-1)^2-5(-1)+6\) |
| \(=\) | \(-1-2+5+6\) | |
| \(=\) | \(8\) |
\(P(1)=0\), so \(z-1\) is a factor; the remainder on dividing by \(z+1\) is \(8\).
Group in pairs, then split the sum of squares over \(\mathbb{C}\):
| \(z^3+2z^2+4z+8\) | \(=\) | \(z^2(z+2)+4(z+2)\) |
| \(=\) | \((z+2)(z^2+4)\) | |
| \(=\) | \((z+2)(z-2i)(z+2i)\) | |
| \(z+2=0\) | \(\Rightarrow\) | \(z=-2\) |
| \(z^2+4=0\) | \(\Rightarrow\) | \(z=\pm 2i\) |
\((z+2)(z-2i)(z+2i)\); zeros \(z=-2,\ \pm 2i\).
The conjugate is a second zero; the pair gives a real quadratic factor:
| \(\text{conjugate zero}\) | \(=\) | \(\overline{2+3i}=2-3i\) |
| \(\text{sum}\) | \(=\) | \((2+3i)+(2-3i)=4\) |
| \(\text{product}\) | \(=\) | \((2+3i)(2-3i)=4+9=13\) |
| \(\text{quadratic factor}\) | \(=\) | \(z^2-4z+13\) |
Write \(P(z)=(z^2-4z+13)(z-c)\) and match the constant term \(13(-c)=13\):
| \(13(-c)\) | \(=\) | \(13\) |
| \(c\) | \(=\) | \(-1\) |
| \(P(z)\) | \(=\) | \((z^2-4z+13)(z+1)\) |
\((z^2-4z+13)(z+1)\); zeros \(z=2\pm 3i,\ -1\).
Treat it as a quadratic in \(z^2\); factorise, then solve each factor over \(\mathbb{C}\):
| \(z^4-z^2-12\) | \(=\) | \((z^2-4)(z^2+3)\) |
| \(z^2-4=0\) | \(\Rightarrow\) | \(z=\pm 2\) |
| \(z^2+3=0\) | \(\Rightarrow\) | \(z^2=-3\) |
| \(z\) | \(=\) | \(\pm\sqrt{3}\,i\) |
\(z=\pm 2,\ \pm\sqrt{3}\,i\).
Common pitfalls
Frequently asked questions
What is the difference between the factor theorem and the remainder theorem?
The remainder theorem says the remainder when \(P(z)\) is divided by \(z-a\) is \(P(a)\). The factor theorem is the case \(P(a)=0\): then the remainder is zero, so \(z-a\) is a factor.
What is the complex conjugate root theorem?
If a polynomial has real coefficients and \(a+bi\) is a zero, then its conjugate \(a-bi\) is also a zero. Non-real zeros of a real polynomial always come in conjugate pairs.
How do you factorise a real cubic given one complex zero?
Take the conjugate as a second zero, multiply the pair to a real quadratic \(z^2-2az+(a^2+b^2)\), then divide it into the cubic (or match constants) to find the remaining linear factor.
What does the fundamental theorem of algebra say?
Every polynomial of degree \(n\ge 1\) has exactly \(n\) zeros over \(\mathbb{C}\), counting multiplicity, so it factors into \(n\) linear factors over the complex numbers.
Does the conjugate root theorem work with imaginary coefficients?
No. It requires real coefficients. If the coefficients are non-real, the conjugate of a zero need not be a zero, so you must factorise directly rather than pairing.
How many roots does a quartic have over the complex numbers?
Exactly four, counting multiplicity. They may be four real, two real and a conjugate pair, or two conjugate pairs, but there are always four in total over \(\mathbb{C}\).