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.
Let $f\in\mathbb{Z}[x]$ be given by
$$
f=x^n-ax^{n-1}-bx-1$$
where $n\ge 3,a\ge 1,b\ge 0$
Claim:$\;f$ has no repeated roots.
Proof:
Suppose otherwise.
Our goal is to derive a contradiction.
First note that if $f$ has a rational root $r$, then
- $r=-1$.$\\[4pt]$
- $n$ is odd.$\\[4pt]$
- $b=a+2$.$\\[4pt]$
- $f'(-1)=(a+1)(n-2) > 0$.
hence no repeated root of $f$ is rational.
Let $w$ be a repeated root of $f$.
As you showed (but with a minor correction to your result), $w$ must satisfy the equation
$$
b(n-1)w^2-\bigl(ab(n-2)-n\bigr)w-a(n-1)=0
$$
so we can't have $b=0$, else $w$ would be rational.
Hence $w$ is a root of the quadratic polynomial
$$
p=x^2-\left(\frac{ab(n-2)-n}{b(n-1)}\right)x-\frac{a}{b}
$$
which must be irreducible in $\mathbb{Q}[x]$ (since $w$ is not rational).
Since $p$ is irreducible in $\mathbb{Q}[x]$ and $w$ is a common root of $p$ and $f$, it follows that $p$ is a factor of $f$ in $\mathbb{Q}[x]$, hence we can write $f=pg$ for some $g\in\mathbb{Q}[x]$.
But $w$ is a simple root of $p$ (since $p$ is irreducible in $\mathbb{Q}[x]$), and by assumption, $w$ is a repeated root of $f$, hence from $f=pg$, we get that $w$ is a root of $g$, so $p$ is a factor of $g$ in $\mathbb{Q}[x]$.
It follows that $p^2$ is a factor of $f$ in $\mathbb{Q}[x]$.
Note that the discriminant of $p$ is positive, so the roots of $p$ are real, and the product of the roots of $p$ is negative, so one of the roots of $p$ is negative, and the other is positive.
But the roots of $p$ are repeated roots of $f$, contradiction, since by Descartes' rule of signs, $f$ can't have a repeated positive root.
Best Answer
An intuitive explanation: if you consider the polynomial $(x-1)(x-1-\varepsilon)$ $(\varepsilon\ne0)$, it has two roots, $1$ and $1+\varepsilon$. When $\varepsilon\to0$, the second root tends to $1$, so we consider that, in the equation $(x-1)^2=0$, the root $1$ counts for two, whence the multiplicity $2$.