*This is a basic article template, excerpted from the (at the time of this writing) current version of the article, Fundamental Theorem of Algebra.*

*Here is the source:*

```
++ The Theorem
``A polynomial of positive degree over the field [[$\mathbb{C}$]] of complex numbers has a root in [[$\mathbb{C}$]].''[((bibcite clark))]
++ Overview
The polynomial [[$x^2 + 1$]] has no ``root'' in real numbers because [[$x^2 + 1 = 0$]] if and only if [[$x^2 = -1$]], and any real number squared equals [[$0$]] or a positive number.
Expanding one’s search to the complex numbers produces an answer, the square root of [[$-1,$]] also known as [[$\mathrm{i}:$]] [[$\mathrm{i}^2 = -1,$]] so [[$\mathrm{i}^2 + 1 = -1 + 1 = 0$]]. The Fundamental Theorem of Algebra says there is always a solution to a polynomial (that may have complex coefficients) if we're allowed to search through the complex numbers too.
A corollary (a relatively simple extension of the theorem) says that if the largest power in the polynomial is [[$n$]], then there are [[$n$]] complex solutions to a polynomial. However, some of the solutions may be repeats.
Clark[((bibcite clark))] cites Ankeny[((bibcite ankeny))] for his proof.
[[toc]]
++ Groundwork
**Notation and terms**
Following Clark's[((bibcite clark))] practice [[$fz$]] is written instead of [[$f(z)$]]. It's not particularly clear to me, however, why he then uses the parenthesized notation [[$\phi(z)$]] and [[$\psi(z)$]], although he does write [[$\phi \alpha$]].
[[math]]
fz = c_0 + c_1 z + \ldots + c_{n-1}z^{n-1} + c_n z^n \text{, where } n \geq 1
[[/math]]
[[math]]
\bar{f}z = \bar{c}_0 + \bar{c}_1 z + \ldots + \bar{c}_{n-1}z^{n-1} + \bar{c}_n z^n \text{, where } n \geq 1
[[/math]]
[[math]]
c = a + b\mathrm{i} \text{ if and only if } \bar{c}=a-b\mathrm{i}
[[/math]]
[[math]]
\phi = f\bar{f} = a_0 + a_1 z + \ldots + a_{2n} z^{2n}
[[/math]]
[[math]]
\phi = az^{2n} - \psi(z).
[[/math]]
[[math]]
-\psi(z) = a_0 + a_1 z + \ldots + a_m z^m, \text{ for } m<2n
[[/math]]
**Restate theorem**
If [[$f$]] is a polynomial of positive degree over the field [[$\mathbb{C}$]] of complex numbers then [[$f$]] has a root in [[$\mathbb{C}$]].
++ Assumption and Goal
**Assume**
[[$f$]] is a polynomial of positive degree over the field [[$\mathbb{C}$]] of complex numbers.
**Deduce**
[[$f$]] has a root in [[$\mathbb{C}$]].
[...]
because
[[math label888]]
\phi(z) = az^{2n} - \psi(z)
[[/math]]
[...]
(See also Eq. ([[eref label888]]).)
[...]
[[math label727]]
\frac{|a_0| + |a_1||z| + \cdots + |a_m||z|^m}{|a||z|^{2n}} \leq \frac{|a_0| + |a_1| + \cdots + |a_m|}{|a||z|^{2n-m}}
[[/math]]
because[[footnote]][...]Hence, comparatively speaking, the numerator is relatively larger than (or equal to) the denominator compared to before the change.[[/footnote]]
[...]
++ Notes
[[footnoteblock title=""]]
++ Sources
[[bibliography title=""]]
: ankeny : Ankeny, N.C. ``One more proof of the fundamental theorem of algebra.'' //Am. Math Monthly//, 54 (1947) 464, cited in Clark.
: clark : Clark, A. //Elements of Abstract Algebra//. New York: Dover, 1984 [orig. 1971].
: dennery : Dennery, P., and A. Krzywicki. //Mathematics for Physicists//. New York: Dover, 1996 [orig. 1967].
: irving : Irving, R.S. //Integers, Polynomials, and Rings//. New York: Springer, 2004.
: shankar : Shankar, R. //Basic Training in Mathematics: A Fitness Program for Science Students//. New York: Plenum, 1995.
[[/bibliography]]
[[include end-material]]
[!-- remove initial space that was inserted so "include" code would show up--]
```

*And here is how it's rendered:*

## The Theorem

“A polynomial of positive degree over the field $\mathbb{C}$ of complex numbers has a root in $\mathbb{C}$.”[2]

## Overview

The polynomial $x^2 + 1$ has no “root” in real numbers because $x^2 + 1 = 0$ if and only if $x^2 = -1$, and any real number squared equals $0$ or a positive number.

Expanding one’s search to the complex numbers produces an answer, the square root of $-1,$ also known as $\mathrm{i}:$ $\mathrm{i}^2 = -1,$ so $\mathrm{i}^2 + 1 = -1 + 1 = 0$. The Fundamental Theorem of Algebra says there is always a solution to a polynomial (that may have complex coefficients) if we're allowed to search through the complex numbers too.

A corollary (a relatively simple extension of the theorem) says that if the largest power in the polynomial is $n$, then there are $n$ complex solutions to a polynomial. However, some of the solutions may be repeats.

Clark[2] cites Ankeny[1] for his proof.

## Groundwork

**Notation and terms**

Following Clark's[2] practice $fz$ is written instead of $f(z)$. It's not particularly clear to me, however, why he then uses the parenthesized notation $\phi(z)$ and $\psi(z)$, although he does write $\phi \alpha$.

(1)**Restate theorem**

If $f$ is a polynomial of positive degree over the field $\mathbb{C}$ of complex numbers then $f$ has a root in $\mathbb{C}$.

## Assumption and Goal

**Assume**

$f$ is a polynomial of positive degree over the field $\mathbb{C}$ of complex numbers.

**Deduce**

$f$ has a root in $\mathbb{C}$.

[…]

because

(7)[…]

(See also Eq. (7).)

[…]

(8)because^{1}

[…]

## Notes

## Sources

*Am. Math Monthly*, 54 (1947) 464, cited in Clark.

*Elements of Abstract Algebra*. New York: Dover, 1984 [orig. 1971].

*Mathematics for Physicists*. New York: Dover, 1996 [orig. 1967].

*Integers, Polynomials, and Rings*. New York: Springer, 2004.

*Basic Training in Mathematics: A Fitness Program for Science Students*. New York: Plenum, 1995.

## Advisory

*Please note that I am not a mathematician and so the presentation of proofs that I make may be deeply flawed. I'm using this writing process to figure out what I'm reading. Please consult more authoritative sources as well.*

Feel free to contact me by leaving a comment or sending me a private message.