Fundamental theorem of algebra and factoring?

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The fundamental theorem of algebra states that every polynomial with complex coefficients has at least one complex root, which is equivalent to the assertion that such polynomials can be expressed as a product of linear factors. This equivalence allows for the iterative process of factoring polynomials by first identifying a root and then applying the theorem to the resulting polynomial of lower degree. The discussion clarifies that the factorization process relies on algebraic identities to derive the factored form from the polynomial. Additionally, a reference to the polynomial remainder theorem is provided for further understanding. Overall, the fundamental theorem of algebra is essential for polynomial factorization in the complex plane.
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Is the fundamental theorem of algebra (for polynomials on the complex plane) equivalent to the statement that any polynomial p of degree n>0 can be written

p(z) = c(z - a_1 ) (z- a_2) \cdot \cdot \cdot (z - a_n )

or am I missing some subtle distinction? And if not equivalent, does the theorem imply this statement?
 
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No, the "fundamental theorem of algebra" (every polynomial equation with complex coefficients has at least one complex root) is exactly equivalent to the statement that every polynomial over the complex numbers can be written as a product of linear factors.
 
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Yes, it is equivalent.
 
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Starting with "fundamental theorem of algebra" (every polynomial equation with complex coefficients has at least one complex root)...say the root is ##a_1## then

##p (a_1) = ca_1^n + c_{n-1} a_1^{n-1} + c_{n-2} a_1^{n-2} + \dots + c_1a_1 + c_0 = 0##.

Write

##p(z) = p(z) - p(a_1) = c (z^n - a_1^n) + c_{n-1} (z^{n-1} - a_1^{n-1}) + c_{n-2} (a_1^{n-2} - z^{n-2}) + \dots + c_1 (z - a_1) \quad Eq1##

and use the algebraic identity:

##z^k - a_1^k = (z - a_1) (z^{k-1} + a_1 z^{k-2} + a_1^2 z^{k-3} + \dots + a_1^{k-2} z + a_1^{k-1})##

to factor out ##(z - a_1)## from every term in Eq1, reducing it to the form:

##p(z) = c (z - a_1) q(z)##

where ##q(z)## is a polynomial of order ##n-1##:

##q(z) = z^{n-1} + d_{n-2} z^{n-2} + \dots + d_1 z + d_0##.

We then apply the fundamental theorem of algebra to ##q(z)## and iterate.
 
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HallsofIvy said:
No, the "fundamental theorem of algebra" (every polynomial equation with complex coefficients has at least one complex root) is exactly equivalent to the statement that every polynomial over the complex numbers can be written as a product of linear factors.
What do you mean? Isn't that what the OP wants to present?
 
Thanks, all. I saw the factored form in a proof without justification and figured it was the Fundamental Theorem.
 
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