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Question: Define the largest natural number m such that the polynomial
$$P(x) = x^5-3x^4+5x^3-7x^2+6x-2$$
be divisible by $(x-1)^m$.
Hint $\ $ If a polynomial $\,f(x)\,$ has power series $\,c_k x^k + \cdots +c_{k+j} x^{k+j},\,\ c_k\ne 0,\,$ then the highest power of $\,x\,$ that divides $\,f(x)\,$ is $\,k,\,$ the order of the power series at $\,x = 0.\,$ An analogous remark holds for divisibility by $\,x-1\,$ using a series at $\,x = 1.\,$ Computing its derivatives then evaluating them at $\,x = 1,\ $ yields $\,\ \color{#0a0}{0 = P(1) = P'(1) = P''(1)},\ $ but $\ \color{#c00}{P'''(1)\ne 0}.$
$$\quad P(x)\, =\, \color{#0a0}{P(1)} + \color{#0a0}{P'(1)}\, (x-1) + \dfrac{\color{#0a0}{P''(1)}}2 (x-1)^2 + \dfrac{\color{#c00}{P'''(1)}}6\, (x-1)^3 + \cdots$$
Therefore, we see that the highest power of $\,x-1\,$ that divides $\,P(x)\,$ is $\,\ldots$
Using the Euclidean division of $x^5-3x^4+5x^3-7x^2+6x-2$ and $x-1$ we get: $$x^5-3x^4+5x^3-7x^2+6x-2=(x^4-2x^3+3x^2-4x+2)(x-1)$$
Then apply the Euclidean division of $x^4-2x^3+3x^2-4x+2$ and $x-1$.
Then we get $x^4-2x^3+3x^2-4x+2=q(x-1)$.
Then apply the Euclidean division of $q$ and $x-1$ and so on.
