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Would someone mind helping me with this question? The more detailed possible so I can have 100% of understanding. Thanks.

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 $(x-1)^m$.

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closed as off-topic by Andrés E. Caicedo, RecklessReckoner, 900 sit-ups a day, Gina, T. Bongers Jul 24 '14 at 23:41

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Run the divison algorithm. – Adam Hughes Jul 24 '14 at 20:34
@AdamHughes: That's not very helpful! – TonyK Jul 24 '14 at 20:45
@TonyK I disagree, if you do division by $(x-1)$ you can go a certain number of times with no remainder, which gives the answer. – Adam Hughes Jul 24 '14 at 20:46
@AdamHughes: That's more like it :-) – TonyK Jul 24 '14 at 20:52
up vote 3 down vote accepted

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 $\,\ldots$

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Thanks! Really helpful! – Vitor Costa Jul 24 '14 at 21:03

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.

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Nice!!!!!!!!!!! – I like Serena Jul 26 '14 at 11:14

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