As per the statement of Fundamental Theorem of Algebra, any polynomial which is non-constant accompanied by complex coefficients consist of not less than 1 complex root.

The theorem also means that any polynomial having complex coefficients belonging to n degree contains n complex roots, which are calculated with multiplicity.

The field F, having the property which every polynomial having coefficients within F contains a root within F, which is known as “**Algebraically Closed**”, therefore the “Fundamental Theorem of Algebra” gives you that statement that:

Complex Numbers, C’s field is **Algebraically Closed. **

**For example:**

*x*²+1 polynomial consists of no real roots, but it has two complex roots *i* and −*i*

*x*²+ i polynomial is having 2 complex roots namely, ±2 / 1−*i*

One may expect that the polynomials accompanied by complex coefficients are having problems with the non-existence of the roots which are identical to real polynomials; i.e. it’s not logical to imagine that a few of polynomials such as

*x*3 + *ix*² − (1+*πi*) *x *−*e*

will not be having the complex root, plus to find such root which needs to look out in a bit larger field that encompasses the field of complex numbers.

As per Fundamental Theorem of Algebra, this is not that very case: all roots of polynomial accompanied by complex coefficients are witnessed existing within the already established field of complex numbers.

## Factoring

This section of the article includes a much more precise statement of the fundamental theorem of algebra’s various equivalent forms. This needs a definition of root’s multiplicity of the polynomial.

**Definition:** Root’s multiplicity ‘r’ of the polynomial *f*(*x*) exists as the largest or the greatest positive integer “*k”* in a way that (x-r)^k divides the function, *f*(*x*).

Similarly, it’s the positive integer which is the smallest one – *k* in a way that f^(k) (r) ≠ 0 where,

*f^*(*k*) represents = *k*th derivative of f.

Assuming F as the field and following are the **equivalents:**

- Every polynomial with a -accompanied with coefficients within F contains root in ‘F”.
- Every polynomial with a non-constant having n degree with coefficients within F contains ‘n’ roots in F which are calculated with multiplicity.
- Every polynomial non-constant value with the coefficients within F entirely splits as linear factors’ product with the coefficients within F.

Evidently, (3)⇒(2)⇒(1), therefore, the exclusive non-trivial section is (1)

To view this, introduce into n’s degree of f(x). The fundamental case of n=1 is clarified.

Now, let us assume the result is for the polynomials of n-1 degree. Accordingly, letting the *f*(*x*) be the polynomial of ‘n’ degree. By (1), now, f(x) contains root *a*.

The argument of the standard division algorithm presents that “x-a” is *f *(*x*)’s factor:

On dividing, *f*(*x*) by *x*−*a* to achieve *f*(*x*)=(*x*−*a*) *q*(*x*) + *r*, where *r* is the constant polynomial.

Filling in ‘*a’* to both of the sides & give a value, 0=(*a*−*a*) *q *(*a*)+*r*, so *r *= 0. So, *f*(*x*)=(*x*−*a*) *q*(*x*). But, *q*(*x*) is a polynomial of degree *n*−1, so it divides into linear factors’ product by an “Inductive Hypothesis” or “Inductive Assumption”. Thus, *f*(*x*) performs as well. Therefore, the result is verified by induction.

The fundamental theorem of algebra states that field C belonging to complex numbers contains property (1), therefore, going by theorem stated , it should have (1), (2), and (3) properties.