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I have this problem,

Let $$p_n(x)=nx^{n+1}-(n+1)x^n+1 \in \mathbb Q[x]$$ for any positive integer $n$. Show that there exist a polynomial $q_n(x) \in \mathbb Q[x] $ satisfying $p_n(x)=(x-1)^2q_n(x)$.

initially I thought that I could use the division algorithm, such that $p_n(x)=(x-1)^2q_n(x)+v(x)$ but is there a reason that we could say that $v(x)=0$?

user26857
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somethin'
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1 Answers1

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Hint: $\;p_n^{\,'}(x)=n(n+1)x^{n-1}(x-1)\,$, so $p_n(1)=p_n^{\,'}(1)=0\,$.

[ EDIT ]  Alternative hint:

$$ \begin{align} p_n(x) & = nx^n(x-1)-(x^n-1) \\ & = nx^n(x-1) - (x-1)(x^{n-1}+x^{n-2}+\cdots+1) \\ & = (x-1)(nx^n-x^{n-1}-x^{n-2}-\cdots-1) \\ & = (x-1)\big((x^n - x^{n-1}) + (x^n-x^{n-2})+ \cdots + (x^n-1)\big) \end{align} $$

Note that each term in the latter sum is divisible by $x-1\,$.

dxiv
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