With complex numbers, every quadratic has two solutions. The fundamental theorem of algebra says the same story holds for every degree, which is why we solve polynomial equations with confidence that the solutions are there to find.
Key idea
A polynomial of degree $n \ge 1$ has exactly $n$ complex zeros, counting multiplicity. If its coefficients are real, its non-real zeros come in conjugate pairs: if $a + bi$ is a zero, so is $a - bi$.
Why it works
The first part is a deep theorem (first proved around 1800); we use it without proof. The pairing is easy to see. Conjugating a sum or product gives the sum or product of the conjugates, and a real number is its own conjugate. So if $p(a + bi) = 0$ and every coefficient is real, conjugating the whole equation gives $p(a - bi) = \overline{0} = 0$. A consequence: a real polynomial of odd degree always has at least one real zero, since the non-real ones pair off.
Worked examples
Worked example: Using a given complex zero
$3 + i$ is a zero of $p(x) = x^3 - 8x^2 + 22x - 20$. Find all the zeros.
- The coefficients are real, so $3 - i$ is a zero too.
- Together: $(x - (3 + i))(x - (3 - i))$ $= (x - 3)^2 - i^2 = x^2 - 6x + 10$.
- Divide: $p(x) = (x^2 - 6x + 10)(x - 2)$.
Answer: $2$, $3 + i$ and $3 - i$
Worked example: Counting real zeros
A degree-4 polynomial with real coefficients has $2i$ as a zero. What could the other zeros be?
- $-2i$ must also be a zero. That uses 2 of the 4.
- The remaining 2 are either both real or another conjugate pair.
Answer: $-2i$, plus two real zeros or another conjugate pair
Common mistake
Forgetting the conjugate. If $1 + 2i$ is a zero of a polynomial with real coefficients, then $1 - 2i$ is a zero too, and together they account for two of the zeros. Leaving it out makes the count wrong and makes it impossible to build a factor with real coefficients.