Fundamental Theorem of Algebra

The fundamental theorem states that a polynomial equation

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where z and the a1 are complex numbers (that includes real numbers), n is an integer > 1, and in which the equation has a solution (also called a root or a zero) in complex numbers. In other words, there always exists a complex number z (which could be a real number) such that P(z) = 0.

Overview

The theorem was stated by Albert Girard in 1629, and the first correct proof was by Argand in 1806. It is unclear who named the theorem. It is possible the name was applied in the early nineteenth century because the solution of polynomial equations was then a frontier of mathematics (proofs by Abel [1823] and Galois [published 1846] that no formula exists for solving fifth- and higher-degree polynomials).

The number n is called the degree of the polynomial. If n = 0, the polynomial is a constant (else the polynomial is unbounded). A function is bounded if it has both a maximum and a minimum value for any valid inputs.

If a solution z0 exists and n > 1, the polynomial can be divided by x - z0 to give a polynomial of degree n – 1. By the fundamental theorem, this new polynomial also has a solution. By repeating this process, it is possible to find n solutions to any nth-degree polynomial equation. But these solutions are not necessarily distinct (for example P(z) = (z – 2)n = 0 has n solutions, all of them equal to 2). Hence the fundamental theorem is sometimes stated as "any polynomial equation of degree n has n solutions, not necessarily distinct."

The fundamental theorem shows that complex numbers are closed under root extraction (let P(z) = znk = 0 (z is the nth root of k). By the theorem there exists a complex number z that solves this equation; that is, k has a nth root.

Proof

Numerous proofs of the fundamental theorem have been developed. Following is an outline of a fairly simple proof that requires Liouville’s theorem from complex analysis: If an entire function (i.e., a function that is defined and differentiable at every point of the complex plane; all polynomials are entire functions) is bounded, then it is a constant function.

Assume there exists a polynomial function P(z) of degree n > 1 that has no solution. Then P(z) must be always positive or always negative. If positive then it has a minimum value P(z0) > 0. Now consider the function 1/P(z), which is entire because P(z) by assumption is never equal to zero. 1/P(z) is bounded above by 1/P(z0) and below by zero. Hence, by Liouville’s theorem, 1/P(z) is constant, and therefore P(z) is constant. (Similar argument if P(z) < 0.) But this contradicts the assumption that P(z) is of degree > 1, because no polynomial of degree > 1 is constant. The contradiction shows that no P(z) can exist without a solution.

Bibliography

Angel, Allen R., and Dennis C. Runde. Elementary Algebra for College Students. Boston: Pearson Prentice, 2011.

Larson, Ron, and David C. Falvo. Precalculus. Boston: Cengage, 2014.

Lial, Margaret L., E J. Hornsby, and Terry McGinnis. Intermediate Algebra. Boston: Pearson, 2012.

Miller, Julie. College Algebra Essentials. New York: McGraw, 2013.

Young, Cynthia Y. College Algebra. Hoboken, NJ: Wiley, 2012.