Fundamental Theorem of Algebra

This theorem bridges the gap between algebra and analysis, assuring us that polynomial equations always have solutions in the complex plane.

Statement

Theorem (Fundamental Theorem of Algebra)

Every non-constant single-variable polynomial with complex coefficients has at least one complex root.

Corollary: A polynomial $P (x)$ of degree $n \geq 1$ has exactly $n$ roots in the complex numbers (counting multiplicity).

P (x) = c (x - r_{1}) (x - r_{2}) \dots (x - r_{n})

If $P (x)$ has real coefficients:

Conjugate Pairs: If $z = a + bi$ is a root, then $\overset{z}{ˉ} = a - bi$ is also a root.
Odd Degree: Every polynomial of odd degree with real coefficients has at least one real root (since complex roots come in pairs).
Factorization: Any real polynomial can be factored into linear terms $(x - r)$ and irreducible quadratic terms $(x^{2} + a x + b)$ over $R$ .

While the theorem guarantees

n

complex roots, finding how many are real is a different challenge.

Let

P (x)

be a polynomial with real coefficients.

The number of positive real roots is equal to the number of sign changes in the sequence of coefficients, or less than that by an even number.
The number of negative real roots is equal to the number of sign changes in the coefficients of $P (- x)$ , or less than that by an even number.

Example

P (x) = x^{3} + x^{2} - x - 1

Sign changes in $P (x)$ : $(+ \to + \to - \to -)$ . One change. $⟹$ Exactly 1 positive real root.
$P (- x) = - x^{3} + x^{2} + x - 1$ .

Sign changes:

(- \to + \to + \to -)

. Two changes.

⟹

2 or 0 negative real roots.

Exercise (Problem 1)

Explain why a polynomial of degree 4 with real coefficients cannot have exactly 3 real roots.

Exercise (Problem 2)

Use Descartes' Rule of Signs to determine the possible number of positive and negative real roots for

f (x) = x^{5} - x^{4} + 3 x^{3} + 9 x^{2} - x + 5

Exercise (Problem 3)

Prove that if a polynomial

P (x)

has real coefficients and

P (i) = 0

, then

x^{2} + 1

is a factor of

P (x)