Fundamental theorem of algebra and factoring?

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  • #1
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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

[tex]p(z) = c(z - a_1 ) (z- a_2) \cdot \cdot \cdot (z - a_n )[/tex]

or am I missing some subtle distinction? And if not equivalent, does the theorem imply this statement?
 

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  • #2
HallsofIvy
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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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  • #3
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Yes, it is equivalent.
 
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  • #4
julian
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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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  • #5
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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.
What do you mean? Isn't that what the OP wants to present?
 
  • #7
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Thanks, all. I saw the factored form in a proof without justification and figured it was the Fundamental Theorem.
 

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