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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.

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  • $\begingroup$ Your 1. is equivalent to every non-zero polynomial has at least one complex root, now use division and induction to get 2. $\endgroup$ – Jonathan Feb 23 '18 at 2:20
  • $\begingroup$ @Jonathan Thank you. I'm curious about these concepts, and it's not an exercise or test. Do you have any more or less accessible reference? $\endgroup$ – Antoni Parellada Feb 23 '18 at 2:22
  • $\begingroup$ en.wikipedia.org/wiki/Algebraically_closed_field $\endgroup$ – fredgoodman Feb 23 '18 at 2:58
  • $\begingroup$ Many HS-level algebra/precalculus textbooks prove The Complete Factorization Theorem, assuming The Fundamental Theorem of Algebra, by repeatedly applying The Factor Theorem. For example, the text "Algebra and Trigonometry, 3rd Ed (2011)", by the Stewart/Redlin/Watson is one such text. $\endgroup$ – quasi Feb 23 '18 at 3:03
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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.

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  • $\begingroup$ Thank you for the comments (partially quoted). I think this is it. If there is a good post online that I can link to the "complete factorization theorem" please edit the post, or send it as a comment. $\endgroup$ – Antoni Parellada Feb 23 '18 at 20:28

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