The Fundamental Theorem of Algebra and Complex Zeros

By the end of this lesson, you will be able to state the Fundamental Theorem of Algebra, find all complex zeros of polynomials, apply the Conjugate Zeros Theorem, and write polynomials with given complex zeros.

The Big Idea

The Fundamental Theorem of Algebra guarantees that every polynomial of degree $n$ with complex coefficients has exactly $n$ zeros (counting multiplicity) in the complex number system. This is profound: a degree-4 polynomial always has exactly 4 zeros — you just may need complex numbers to find them all.

Fundamental Theorem of Algebra

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

Corollary: A degree-$n$ polynomial has exactly $n$ zeros (counting multiplicity) in $\mathbb{C}$.

Real corollary: A degree-$n$ polynomial with real coefficients has at most $n$ real zeros, but always exactly $n$ complex zeros.

Conjugate Zeros Theorem

If $p(x)$ has real coefficients and $a + bi$ (with $b \neq 0$) is a zero, then its complex conjugate $a - bi$ is also a zero.

Complex zeros of real-coefficient polynomials always come in conjugate pairs.

Example 1 — Finding All Zeros

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

Factor: $x(x^3 - x^2 + x - 1) = x[x^2(x-1) + (x-1)] = x(x-1)(x^2+1)$

Zeros: $x = 0$, $x = 1$, $x^2 + 1 = 0 \implies x = \pm i$

All four zeros: $0, 1, i, -i$ (conjugate pair $\pm i$) ✓

Example 2 — Building a Polynomial from Zeros

Write a polynomial with real coefficients and zeros $3, 2 - i$.

By Conjugate Zeros Theorem, $2 + i$ must also be a zero.

$(x-3)[x-(2-i)][x-(2+i)] = (x-3)[(x-2)^2+1] = (x-3)(x^2-4x+5)$

$= x^3 - 4x^2 + 5x - 3x^2 + 12x - 15 = x^3 - 7x^2 + 17x - 15$

Example 3 — Given a Complex Zero, Find the Rest

$p(x) = x^3 - 2x^2 + 4x - 8$ has a zero at $2i$. Find all zeros.

By conjugate theorem: $-2i$ is also a zero.

Factor: $(x - 2i)(x + 2i) = x^2 + 4$

Divide: $p(x) = (x^2+4)(x-2)$

All zeros: $2, 2i, -2i$

Counting Zeros

Type