Math Core

A linear equation has one solution, and a quadratic has two once you allow complex numbers. Does that pattern continue? It does, and the result is important enough to be called the fundamental theorem of algebra. It tells you exactly how many zeros to look for, so you know when you've found them all.

How many zeros?

Fundamental theorem of algebra

Every polynomial of degree $n \ge 1$ with complex coefficients has at least one complex zero.

As a consequence, a polynomial of degree $n$ has exactly $n$ complex zeros, counted with multiplicity, and it factors completely into $n$ linear factors:

P(x) = a(x - r_1)(x - r_2)\cdots(x - r_n).

Why does "at least one" lead to "exactly $n$ "? If $r_1$ is a zero, the factor theorem gives $P(x) = (x - r_1)Q(x)$ , where $Q$ has degree $n - 1$ . The theorem applies again to $Q$ , so it has a zero $r_2$ , and so on. After $n$ steps you've split off $n$ linear factors and only a constant $a$ (the leading coefficient) is left.

Remember that real numbers count as complex numbers: $5 = 5 + 0i$ . "Complex zeros" includes the real ones.

Multiplicity

Some zeros repeat. In $P(x) = (x - 2)^3(x + 1)$ , the factor $x - 2$ appears three times.

Definition

Multiplicity

If $(x - r)^k$ is a factor of $P(x)$ but $(x - r)^{k+1}$ is not, then $r$ is a zero of multiplicity $k$ . A zero of multiplicity $2$ is called a double zero.

"Counted with multiplicity" means a double zero counts twice, a triple zero three times, and so on. $P(x) = (x - 2)^3(x + 1)$ has degree $4$ and only two distinct zeros, but $3 + 1 = 4$ when you count multiplicities. The count always matches the degree.

Worked example: Counting zeros

List the zeros of $P(x) = (x - 2)^3(x^2 + 9)$ with their multiplicities.

The degree is $3 + 2 = 5$ , so there are $5$ zeros counted with multiplicity. The factor $(x - 2)^3$ gives the zero $2$ with multiplicity $3$ . For $x^2 + 9 = 0$ , $x^2 = -9$ and $x = \pm 3i$ , each with multiplicity $1$ .

Check the count: $3 + 1 + 1 = 5$ . ✓ So $P(x) = (x - 2)^3(x - 3i)(x + 3i)$ .

Complex zeros come in pairs

When you solved quadratics with negative discriminants, the solutions always came as a pair $a + bi$ and $a - bi$ . That's no accident.

Complex conjugate root theorem

If a polynomial has real coefficients and $a + bi$ (with $b \ne 0$ ) is a zero, then its conjugate $a - bi$ is also a zero, with the same multiplicity.

Two consequences:

A polynomial with real coefficients has an even number of non-real zeros. So a cubic with real coefficients always has at least one real zero, and a quintic does too.
Each conjugate pair multiplies to a quadratic with real coefficients: $(x - (a + bi))(x - (a - bi)) = x^2 - 2ax + (a^2 + b^2)$ .

There's a similar pairing for irrational zeros: if a polynomial has rational coefficients and $a + \sqrt{b}$ is a zero (with $\sqrt{b}$ irrational), then $a - \sqrt{b}$ is a zero too.

Common mistake

The conjugate theorem needs real coefficients. The polynomial $x - i$ has the zero $i$ but not $-i$ . On tests, the problem will say "real coefficients" when it wants you to use the theorem, so watch for that phrase.

Finding all the zeros

Worked example: Real and non-real zeros

Find all the zeros of $P(x) = x^3 - 3x^2 + 4x - 12$ .

Group the terms:

x^2(x - 3) + 4(x - 3) = (x - 3)(x^2 + 4).

So $x = 3$ , or $x^2 = -4$ , which gives $x = \pm 2i$ . The three zeros are $3$ , $2i$ and $-2i$ : one real zero and one conjugate pair, just as the theorems predict for a cubic with real coefficients.

Worked example: Using a known complex zero

Given that $1 + i$ is a zero of $P(x) = x^4 - 2x^3 + 3x^2 - 2x + 2$ , find all its zeros.

The coefficients are real, so $1 - i$ is also a zero. Their factors multiply to a real quadratic:

(x - (1 + i))(x - (1 - i)) = x^2 - 2x + (1^2 + 1^2) = x^2 - 2x + 2.

Divide $P(x)$ by $x^2 - 2x + 2$ using long division:

step	divide	multiply by $x^2 - 2x + 2$	subtract, what's left
1	$x^4 \div x^2 = x^2$	$x^4 - 2x^3 + 2x^2$	$x^2 - 2x + 2$
2	$x^2 \div x^2 = 1$	$x^2 - 2x + 2$	$0$

So $P(x) = (x^2 - 2x + 2)(x^2 + 1)$ . From $x^2 + 1 = 0$ you get $x = \pm i$ . The four zeros are $1 + i$ , $1 - i$ , $i$ and $-i$ .

Building a polynomial from its zeros

The factored form runs in reverse too: if you know the zeros, you can write the polynomial.

Worked example: Lowest degree with real coefficients

Write a polynomial of least degree with real coefficients and leading coefficient $1$ that has zeros $1$ and $2 + i$ .

Because the coefficients are real, $2 - i$ must also be a zero. That makes three zeros, so the least degree is $3$ . Multiply the conjugate pair first:

(x - (2 + i))(x - (2 - i)) = x^2 - 4x + (4 + 1) = x^2 - 4x + 5.

Then multiply by $x - 1$ :

(x - 1)(x^2 - 4x + 5) = x^3 - 4x^2 + 5x - x^2 + 4x - 5 = x^3 - 5x^2 + 9x - 5.

Check: every coefficient is real, as required.

Tip

Multiply conjugate pairs together first. The $i$ terms cancel immediately and you're left working with real numbers. The shortcut $x^2 - 2ax + (a^2 + b^2)$ for the pair $a \pm bi$ saves time.