If $x - r$ is a factor of the polynomial

$$f(x) = a_n x^n + \ldots + a_0,$$

repeated $m$ times $(1 < m \leq n)$, then $r$ is a root of $f'(x) = 0$ repeated $m$ times.

Key concepts you must know

1. Multiplicity (order) of a root
   If $(x-r)^m$ divides $f(x)$ but $(x-r)^{m+1}$ does not, then we say $r$ is a root of multiplicity $m$.

2. Product rule for derivatives
   If $u(x)$ and $v(x)$ are functions,
   $\big(u\,v\big)' = u'v + uv'.$

3. Why one power drops after differentiating
   Differentiating $(x-r)^m$ gives
   $\frac{d}{dx}(x-r)^m = m(x-r)^{m-1},$
   which clearly contains only $(x-r)^{m-1}$.

Logical chain of thought to tackle the proof

1. Factor the polynomial completely:
   $f(x) = (x-r)^m g(x), \qquad g(r)\neq0.$
2. Differentiate using product rule:
   $f'(x) = \big((x-r)^m\big)'g(x) + (x-r)^m g'(x).$
3. Compute each derivative:
   $\big((x-r)^m\big)' = m(x-r)^{m-1},$
   so
   $f'(x) = m(x-r)^{m-1}g(x) + (x-r)^m g'(x).$
4. Factor out the highest common power of $(x-r)$:
   $f'(x) = (x-r)^{m-1}\Big[\,m\,g(x) + (x-r)g'(x)\Big].$
5. Check the bracket at $x=r$:
   At $x=r$, the term $(x-r)$ inside the big bracket is $0$, so only the $m\,g(r)$ part remains, and $g(r)\neq0$. Therefore the bracket is not zero at $x=r$.
6. Conclusion: $(x-r)^{m-1}$ divides $f'(x)$ but $(x-r)^m$ does not.
   Hence $r$ is a root of $f'(x)=0$ of multiplicity $m-1$.

So the statement in the prompt is slightly inaccurate; the correct multiplicity is $m-1$, not $m$ (for $m>1$).

What is the question?

We have a polynomial $f(x)$ and it contains the same bracket $(x-r)$ many times.
If it appears $m$ times, we write
$f(x) = (x-r)^m \times g(x)$
where $g(x)$ is just the ‘left-over’ part (and $g(r)\neq0$ because all the $(x-r)$ factors are already shown).

Now the question says “then $r$ is a root of $f'(x)=0$ repeated $m$ times.”
In plain words: "Does $(x-r)^m$ also divide the derivative?"

Big idea (child-friendly)

Imagine $f(x)$ is like a stack of $m$ identical Lego blocks labelled $(x-r)$.
When you take the slope (derivative), one Lego block always breaks off, so you usually end up with $m-1$ blocks of $(x-r)$ in $f'(x)$.
So $r$ is a root of the derivative, but only $m-1$ times, not $m$ times.

Complete Proof (step-by-step, basic English)

We start with the given information:

$f(x)=a_nx^n+\dots+a_0$

and $(x-r)$ is repeated $m$ times as a factor.

So we can write

$f(x)=(x-r)^m g(x), \; \text{where } g(x) \text{ is a polynomial and } g(r)\neq0.$

1. Differentiate using the product rule:

2. Factor out the common $(x-r)^{m-1}$:

3. Check the bracket at $x=r$:

$\Big[\, m\,g(r) + (r-r)g'(r) \Big]=m\,g(r) \neq 0\quad \text{(because } g(r)\neq0\text{)}.$

Therefore $(x-r)^{m-1}$ is the highest power of $(x-r)$ dividing $f'(x)$.

4. Final result: $r$ is a root of $f'(x)=0$ of multiplicity $m-1$.

So the correct statement is:

If $(x-r)$ appears $m$ times in $f(x)$ with $m>1$, then $(x-r)$ appears $m-1$ times in $f'(x)$.

(Only when $m=1$ does the root generally disappear entirely in the derivative.)