>Equations>What Is the Fundamental Theorem of Algebra?

What Is the Fundamental Theorem of Algebra?

The mathematical theory regarding complex numbers culminates in the fundamental theorem of algebra. As the name suggests, the fundamental theorem of algebra is an important pillar of mathematics. The reason for this is that the theorem guarantees that all algebraic equations with complex coefficients have solutions. An algebraic equation is an equation in the form $f (x) = g (x)$ , where $f (x)$ and $g (x)$ are polynomials.

Rule

The Fundamental Theorem of Algebra

Let $f_{n} (z)$ be a complex polynomial

c_{n} z^{n} + c_{n - 1} z^{n - 1} + \dots + c_{1} z + c_{0}

of degree $n$ with coefficients $c_{n}, c_{n - 1}, \dots, c_{1}, c_{0} \in ℂ$ . Then there exist complex numbers $r_{1}, r_{2}, \dots, r_{n} \in ℂ$ such that

f_{n} (z) = c_{n} (z - r_{1}) \dots (z - r_{n}) .

f_{n} (z) = c_{n} (z - r_{1}) (z - r_{2}) \dots (z - r_{n}) .

The fundamental theorem of algebra states that $f_{n} (z)$ has $n$ uniquely determined zeros $r_{1}, r_{2}, \dots r_{n} \in ℂ$ . The zeros of $f_{n} (z)$ are also called the roots of $f_{n} (z)$ . This meaning of the word “root” is not to be confused with $n$ th roots of a complex number. Even though $f_{n} (z)$ has $n$ uniquely determined roots, that doesn’t mean that all the zeros of $f_{n} (z)$ are unique.

The roots of $f_{n} (z)$ that appear more than once have a multiplicity greater than 1. The multiplicity of a root $n$ is a measure of how many times $(z - r)$ can be factored out of $f_{n} (z)$ . Thus, the correct interpretation of the fundamental theorem of algebra is that $f_{n} (z)$ has $n$ different roots counted with multiplicity. The fundamental theorem of algebra ensures that all polynomials of degree $n$ can be factorized into $n$ linear complex factors.

Proof of the Fundamental Theorem for Quadratic Polynomials

For a quadratic polynomial $f_{2} (z) = a z^{2} + b z + c$ , the fundamental theorem of algebra says that $f_{2} (z)$ can be factorized and written in the form $f_{2} (z) = (z - r_{1}) (z - r_{2})$ , where $r_{1}$ and $r_{2}$ are the roots of $f_{2} (z)$ . You can find the roots of $f_{2} (z)$ by using the quadratic formula

z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

Regardless of whether the discriminant is positive or negative, the quadratic formula yields two distinct roots, so we don’t need to consider those cases. However, if the discriminant equals exactly $0$ , the quadratic formula yields only one solution. To prove the fundamental theorem of algebra for quadratic polynomials, you need to show that if $f_{2} (z)$ has one root $r$ , then $f_{2} (z)$ can be divided twice by $(z - r)$ . This would imply that all quadratic polynomials can be factorized in the way the fundamental theorem describes.

You can find the solutions to the equation $f_{2} (z) = 0$ by using the quadratic formula. This means that you can manipulate the quadratic formula to recover the expression $f_{2} (z) = 0$ by isolating zero on one side of the equation

z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

Adding $\frac{b}{2 a}$ to both sides:

z + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} .

Squaring both sides:

{(z + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}} .

Subtracting $\frac{b^{2} - 4 a c}{4 a^{2}}$ from both sides:

{(z + \frac{b}{2 a})}^{2} - \frac{b^{2} - 4 a c}{4 a^{2}} = 0 .

You can now write $f_{2} (z)$ on the form

f_{2} (z) = {(z + \frac{b}{2 a})}^{2} - \frac{b^{2} - 4 a c}{4 a^{2}} .

When $b^{2} - 4 a c = 0$ , the quadratic polynomial becomes

\begin{array}{l} f_{2} (z) & = {(z + \frac{b}{2 a})}^{2} - \frac{b^{2} - 4 a c}{4 a^{2}} \\ = {(z + \frac{b}{2 a})}^{2} . \end{array}

The root of this polynomial is $r = - \frac{b}{2 a}$ . Just as the fundamental theorem of algebra says, the factor involving this root occur twice in the factorization of $f_{2} (z)$ : $\begin{array}{l} f_{2} (z) & = {(z + \frac{b}{2 a})}^{2} \\ = (z + \frac{b}{2 a}) (z + \frac{b}{2 a}) . \end{array}$

Since the root occurs twice in the factorization, it is called a double root of $f_{2} (z)$ . All quadratic polynomials can be written as a product of two linear factors. This also holds when the equation $f_{2} (z) = 0$ only has one solution. And thus, the fundamental theorem of algebra for quadratic equations has been proven.

Q.E.D

Note! In order to prove the fundamental theorem of algebra in general, considerably more advanced mathematics is required.

Want to know more?Sign UpIt's free!

How to Factorize Complex Polynomials

Go back