Suppose a polynomial equation is pulled out of the air—perhaps this one:
[beautiful math coming... please be patient]
$$7x^6  \frac 12x^4  3 = 2x^9 + 5x$$
Is it guaranteed to have a solution?
In other words, must there exist a value of $\,x\,$ for which it is true?
Alternatively, rearrange the equation above by getting a zero on the righthand side.
Take the resulting expression on the left of the equation, and use it to define a function $\,f\,$:
[beautiful math coming... please be patient]
$$f(x) = 7x^6  \frac 12x^4  3  2x^9  5x$$
Is this function $\,f\,$ guaranteed to have a zero?
In other words, must there exist a value of $\,x\,$ for which $\,f(x)\,$ is zero?
These are precisely the questions for which the Fundamental Theorem of Algebra
provides a beautiful answer!
Does a polynomial equation always have a solution?
Short Answer: It depends! What kind of solutions are you looking for?
If you're looking specifically for real number solutions,
then the answer to the question ‘Does a polynomial equation always have a solution?’ is NO.
For example, the polynomial equation $\,x^2 = 1\,$ doesn't have any real number solutions.
Why not? Because every real number, when squared, is greater than or
equal to zero (hence can't possibly equal $\,1\,$).
If you're looking for complex number solutions (which include any real number solutions), then you're in luck.
Indeed, the Fundamental Theorem of Algebra tells us that there always exists a solution,
as long as you look in the set of complex numbers.
(By the way, both $\,i\,$ and $\,i\,$ (where $\,i = \sqrt{1}\,$) are complex number solutions of the equation $\,x^2 = 1\,$.)
The Fundamental Theorem of Algebra
The statement of the Fundamental Theorem of Algebra is short and simple.
Don't let its simplicity fool you—it is a very powerful result.
Recall that the symbol $\Bbb C\,$ (blackboard bold C) represents the set of complex numbers,
and the symbol $\Bbb R\,$ (blackboard bold R) represents the set of real numbers.
THEOREM
the Fundamental Theorem of Algebra
Every nonconstant polynomial with complex coefficients has at least one zero in $\,\Bbb C\,$.
Comments on the Fundamental Theorem of Algebra:

A ‘constant polynomial’ is sonamed because its output never changes—it is constant.
For example, the function $\,f\,$ defined by $\,f(x) = 5\,$ is a constant polynomial—no matter what the input is, the output is always $\,5\,$.

Most constant polynomials have no zeros.
For example, consider $\,f(x) = 5\,$.
Is $\,f(x)\,$ ever equal to zero? No—it's always $\,5\,$.

On the other hand, the constant function $\,g\,$ defined by $\,g(x) = 0\,$ (the zero function) is always zero.
It has infinitely many zeros.

For these reasons, the Fundamental Theorem of Algebra only deals with nonconstant polynomials.

Remember that the set of complex numbers includes the real numbers.
In other words, if $\,x\in\Bbb R\,$, then $\,x\in\Bbb C\,$.
That is, every real number is also a complex number.
Indeed, the real numbers are the $\,x\,$axis (the real axis) in the complex plane.

There are loads of complex numbers that are not real numbers.
Any number that doesn't lie on the real axis in the complex plane isn't a real number.

A ‘polynomial with complex coefficients’ allows
terms of the form $\,ax^n\,$, where $\,a\in\Bbb C\,$.

Is a polynomial like $\,P(x) = 5x^3  2x + 1\,$ a
‘polynomial with complex coefficients’?
YES! Every real number is a complex number.
The Fundamental Theorem of Algebra covers (in particular) all polynomials with real coefficients.

Recall that ‘at least one’ means greater than
or equal to one.

The Fundamental Theorem of Algebra is an existence theorem.
It guarantees that there exists a complex number $\,c\,$ for which $P(c) = 0$,
as long as the function $\,P\,$ is a nonconstant polynomial with complex coefficients.
COROLLARY
to the Fundamental Theorem of Algebra
Let $\,n\,$ be a positive integer.
Every polynomial of degree $\,n\,$ with complex coefficients has
exactly
$\,n\,$ zeros in $\,\Bbb C\,$,
counting
multiplicities.

Recall that a corollary is an interesting consequence of a theorem—usually
something that is easy to prove.
 The positive integers are: $1, 2, 3, \ldots$

Consider the polynomial $\,P\,$ defined by $\,P(x) = (x1)^2(x5)^4(x+9)\,$.
It has degree $\,7\,$—if it were multiplied out, the highest power on $\,x\,$ would be $\,7\,$.
How many zeros does it have?
This question is a bit ambiguous, so needs some clarification, as follows:

The polynomial $\,P\,$ has three distinct (that is, different) zeros: $\,1\,$, $\,5\,$, and $\,9\,$.
If you look at the graph of $\,P\,$, it crosses the $\,x\,$axis at exactly these three places.

The polynomial $\,P\,$ has seven zeros, counting multiplicities:
 the zero $\,1\,$ has multiplicity $\,2\,$ (from the two factors of $\,(x1)\,$, written as $\,(x1)^2\,\ $)
 the zero $\,5\,$ has multiplicity $\,4\,$ (from the four factors of $\,x5\,$, written as $\,(x5)^4\,\ $)
 the zero $\,9\,$ is a simple zero (multiplicity $\,1\,$) (from the factor $\,(x+9)\,\ $)
Counting multiplicities, the seven zeros of $\,P\,$ are: $\,1\,$, $\,1\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,9\,$

‘Counting multiplicities’ means that if $\,c\,$ is a zero of multiplicity $\,k\,$,
then $\,c\,$ is counted as a zero $\,k\,$ times.
How is this corollary an easy consequence of the Fundamental Theorem of Algebra?
Here's the idea:

Start with a polynomial with complex coefficients that has degree $\,n\,$.
For convenience, call this polynomial $\,P\,$ and let $\,z\,$ represent a typical input.

By the Fundamental Theorem of Algebra, $\,P\,$ has a zero in $\,\Bbb C\,$.
Call this zero $\,z_1\,$.

Since $\,z_1\,$ is a zero of $\,P\,$, $\,z  z_1\,$ is a factor of $\,P(z)\,$.
Thus, $\,P(z) = (zz_1)P_1(z)\,$, where $\,P_1\,$ is a polynomial with complex coefficients that has degree $\,n1\,$.

By the Fundamental Theorem of Algebra, $\,P_1\,$ has a zero in $\,\Bbb C\,$.
Call this zero $\,z_2\,$.

Since $\,z_2\,$ is a zero of $\,P_1\,$, $\,z  z_2\,$ is a factor of $\,P_1(z)\,$.
Thus, $\,P_1(z) = (zz_2)P_2(z)\,$, where $\,P_2\,$ is a polynomial with complex coefficients that has degree $\,n2\,$.
Combining results thus far,
$\,P(z) = (z  z_1)(z  z_2)P_2(z)\,$.

Repeat as needed.
Eventually, we have
$\,P(z) = (z  z_1)(z  z_2)\cdots(z  z_n)P_n(z)\,$,
where $\,P_n(z)\,$ is a polynomial with complex coefficients that has degree $\,n  n = 0\,$.
 Since $\,P_n(z)\,$ has degree zero, it is a constant; call it $K$ for simplicity.

Thus,
$\,P(z) = K(z  z_1)(z  z_2)\cdots(z  z_n)\,$ for some complex constant $\,K\,$.

Thus, $\,P\,$ has the $\,n\,$ zeros $\,z_1\,,\,z_2\,,\,\ldots \,,\,z_n\,$.
Some of the zeros $\,z_i\,$ may be identical, which is why we need to count
multiplicities.
 Voila!