Fundamental theorem of algebra - connecting equivalent formulations

Question

How can these two statements of the fundamental theorem of algebra be reconciled into equivalent formulations of the same concept:

1. The field of complex numbers is algebraically closed.

2. Every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.

Answer 1

Fundamental theorem of algebra: The field of complex numbers is algebraically closed = Every non-zero polynomial has at least one complex root.

The number of complex (including real) roots equals the degree of the polynomial. Hence it can be factored into as many linear factors as the degree of the polynomial...

Complete factorization theorem:

A polynomial $P(x)=\sum_{i=0}^n a_ix^i$ with complex coefficients and complex roots $r_1,r_2,\dots,r_n$ can be factored into

$$P(x)=c\cdot (x-r_1)\cdot(x-r_2)\cdots(x-r_n)$$

with $c$ corresponding to a (non-zero) complex number.

Factor theorem:

The factor theorem states that a polynomial has a factor $x - r_1$ if and only if $P(r_1)=0$ (i.e. $r_1$ is a root).

Hence, $P(x)$ can be factored as

$$P(x)=(x-r_1)\cdot Q_1(x)$$

The quotient group $Q_1(x)$ is of degree $n-1.$

"by repeatedly applying The Factor Theorem..."

If $r_2$ is a factor of $Q_1(x),$ then $P(x)$ can be factored into

$$P(x)=(x-r_1)\cdot(x-r_2)\cdot Q_2(x)$$

reducing by one the degree of the quotient polynomial. And this process can be repeated until we arrive at a constant term, $c:$

$$P(x)=c\cdot (x-r_1)\cdot(x-r_2)\cdots(x-r_n).$$

This proves the second formulation of the FTA in the OP:

Every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.