The Fundamental Theorem of Algebra

# The Fundamental Theorem of Algebra

Recall from the Properties of Polynomials page that a polynomial is a function in the form $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$, and that if $a_n \neq 0$ then $\mathrm{deg} (p) = n$. We will now look at some more theorems regarding polynomials, the first of which is extremely important and is known as The Fundamental Theorem of Algebra.

 Theorem 1 (The Fundamental Theorem of Algebra): If $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is a polynomial with $\mathrm{deg} (p) = n$ and where $a_0, a_1, ..., a_n \in \mathbb{C}$ are complex coefficients of $p$, then $p$ has exactly $n$ roots, $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbb{C}$.

We will not prove The Fundamental Theorem of Algebra, however we should note that $\mathbb{R} \subset \mathbb{C}$, and as a result, the coefficients $a_0, a_1, ..., a_n \in \mathbb{C}$ may in fact be real numbers themselves as every real number is a complex number in the form $a + 0i$ where $a \in \mathbb{R}$.

For example, the polynomial $p(x) = x^2 - 9$ is of degree $2$, and so by The Fundamental Theorem of Algebra, this polynomial has $2$ roots. We can easily find this roots by factoring this polynomial as $p(x) = (x - 3)(x + 3)$, and so $\lambda_1 = 3$ and $\lambda_2 = -3$ are the roots to $p$.

Another example is the polynomial $q(x) = x^2 + x + 1$. Once again, this polynomial is of degree $2$ and so by The Fundamental Theorem of Algebra, this polynomial has $2$ roots. This polynomial cannot be factored nicely, so we must use the quadratic equation to find these roots, which is $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$ is the coefficient on the $x^2$ term, $b$ is the coefficient on the $x$ term, and $c$ is the coefficient on the constant term. Therefore we have that,

(1)
\begin{align} \quad \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i \end{align}

Therefore $\lambda_1 = \frac{-1}{2} + \frac{\sqrt{3}}{2} i$ and $\lambda_2 = \frac{-1}{2} - \frac{\sqrt{3}}{2} i$ are the two roots to $q$. Notice that in this example, neither $\lambda_1$ or $\lambda_2$ were real.