# Polynomial Division - “Define the largest natural number…” [closed]

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 $(x-1)^m$.

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, RecklessReckoner, Community, Gina, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 $\,\ldots$

-
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.

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