# Fundamental theorem of algebra and factoring?

1. Jul 23, 2015

### pellman

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?

2. Jul 23, 2015

### HallsofIvy

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.

3. Jul 23, 2015

### micromass

Yes, it is equivalent.

4. Jul 23, 2015

### julian

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.

Last edited: Jul 23, 2015
5. Jul 23, 2015

### tommyxu3

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

6. Jul 23, 2015

### homeomorphic

7. Jul 23, 2015

### pellman

Thanks, all. I saw the factored form in a proof without justification and figured it was the Fundamental Theorem.