- #1

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[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?

- Thread starter pellman
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- #1

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[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?

- #2

HallsofIvy

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- #3

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Yes, it is equivalent.

- #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.

##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.

Last edited:

- #5

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What do you mean? Isn't that what the OP wants to present?

- #6

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https://en.wikipedia.org/wiki/Polynomial_remainder_theorem

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