The fundamental theorem of algebra is stated in the complex domain ($\mathbb C$). This reminded us that maybe we could have an answer with complex ($\mathbb C$) viewpoint.
Let $f(z)$ be a polynomial over $\mathbb C$, and $u\in\mathbb C$ be a root of $f(z)$ of multiplicity $m$. This is expressed by an equation: $$f(z)=(z-u)^m g(z)$$ where $g(z)$ is a polynomial that does not have $u$ as its root.
Clearly $f$ brings $u$ to $0$, from one complex plane to another. This characterizes the notion of root geometrically:
Since $f$ is continuous, points near $u$ are mapped to some points near $0$. By complex analysis, a small loop around $u$ should be mapped to another loop winding $m$ times around the origin. The following picture shows the case when $m=2$:
Example:
For every complex number $a$ we can assign a color for it according to its angle $\operatorname{arg} a$. The brighter color means that the angle is closer to $0^\circ$:
Let $f(z)=(z+1)^3(z-2)$. For every point $z$ on the complex plane we paint the corresponding color for $\operatorname{arg}f(z)$:
The horizontal axis is real and the vertical one is imaginary. The small green squares indicate the unit length.
When we travel along a path $\gamma$ around $z=-1$, which is the root of multiplicity $3$, the color changes (white to black) three times. This shows that the corresponding path $f(\gamma)$ loops three times around the origin of the codomain. The discussion of the other root $z=2$ is similar.
This characterizes the notion of multiplicity geometrically, so I will try to prove this property (or, at least, give an idea of how it happens, because I'm not really familiar with complex analysis). But so far you can just think that the multiplicity is (or coincides with) how many times the color changes. That's the viewpoint that I wanted to provide.
(In fact, you can count the change of the color around a circle that contains all roots. The result coincides with the degree of $f$. This is related to a proof of the fundamental theorem of algebra.)
Proof (or explanation):
I will work on this statement:
Let $f$ has a root $u$ of multiplicity $m$, and $f$ is described as $f(z)=(z-u)^mg(z)$ where $g(z)$ is nonzero at $u$.
Let $\gamma:[0,2\pi]\to\mathbb C$ defined by $\theta\mapsto u+e^{i\theta}$ be a small circle that loops once around $u$ and does not contain other roots inside it. Then $\gamma$ is mapped to a curve $\gamma_2$ by $f$: $$\begin{matrix}\gamma_2:=f\circ\gamma:&[0,2\pi]&\to&\mathbb C\\
& \theta &\mapsto & f(u+re^{i\theta}).\end{matrix}$$ which loops $m$ times around the origin, just as the second picture of this answer.
Since $\gamma$ does not contain other root of $f$, $\gamma$ contains no root of $g$. This means if we let the radius $r$ tend to zero, then $\gamma$ shrinks to $u$ gradually without passing through any root of $g$. Thus the corresponding curve $g\circ \gamma$ can shrink to a point without passing through zero.
(I implicitly used the continuity of $g$.)
Now, let's simplify $\gamma_2$: $$\begin{aligned}
\gamma_2(\theta)=f(u+re^{i\theta}) &= ((u+re^{i\theta})-u)^mg(u+re^{i\theta})\\[0.7em]
&= re^{im\theta}g(u+re^{i\theta}).
\end{aligned}$$
Since $g(u+re^{i\theta})$ can shrink continuously to a point without passing through the origin, the net change of angle of the loop $g(u+re^{i\theta})$ w.r.t the origin is zero.
(I implicitly use some property of the homotopy)
So the net change of angle of $\gamma_2$ is only caused by $re^{im\theta}$, which winds $m$ times around the origin.
The fundamental theorem of algebra is exactly as difficult for real vs. complex coefficients. The reason is that if $f(x) = f_0 + \dots + f_n x^n$ is any polynomial with complex coefficients, then the polynomial
$$\overline{f(\overline{x})} f(x) = (\overline{f_0} + \dots + \overline{f_n} x^n)(f_0 + \dots + f_n x^n)$$
has real coefficients (because it is invariant under coefficient-wise complex conjugation), and its roots are exactly the roots of $f$ together with their complex conjugates (this is a nice exercise). So any proof of the FTA for real coefficients immediately yields the FTA for complex coefficients.
It speaks well of you that you want to understand this result instead of taking it on faith; unfortunately it is one of the first "genuinely difficult" results one comes across in mathematics. There are no really easy proofs, one has to really understand something. You can find a lovely collection of proofs here on MO but all their prerequisites involve a certain amount of analysis or topology or in one case Galois theory and this is unavoidable. (The Galois theory is used to reduce the amount of analysis necessary to the fact that a real polynomial of odd degree has a real root (edit: and, as Daniel Schepler points out, to the fact that a non-negative real number has a real square root), which follows from the intermediate value theorem.)
FWIW I think this proof shared by Kevin McGerty is the least technical on the list but it still requires some familiarity with analysis.
Best Answer
There is no barrier to considering polynomials with complex coefficients and complex roots. The fundamental theorem of algebra is an assertion about those polynomials - each one factors into a product of linear factors.
When the coefficients happen to be real the roots must occur in conjugate pairs.
Edit in response to comment.
If the roots occur in conjugate pairs then the coefficients are real, because $(x- r)(x-\bar r)$ has real coefficients. But the roots must pair up. The polynomial $$ (x-i)^2(x+i) $$ has $i$ and $-i$ as roots but nonreal coefficients.