Prove that 1 is a triple root of a polynomial I'm studying for an exam and trying to prove whether 1 is a triple root for the polynomial:
$$x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$$
for every $n\geq1$.
In our math class we never solved such a problem. So far we only used horner's scheme to prove that someone is a root, double root or triple root.
Can you please help me solve this problem?
 A: Assume that $a$ is a $k$-ary root of some polynomial $p \in \Bbb R[x]$. Then you might write $p$ as $p=(x-a)^kq$ for some polynomial $q$ where $q(a) \neq 0$. Since $p$ (as a function on $\Bbb R$) is differentiable we may differentiate it and find
$$
\mathrm{D}p(x) = k(x-a)^{k-1}q(x) + (x-a)^k \mathrm{D}q(x) = (x-a)^{k-1}(kq(x) + (x-a)\mathrm{D}q(x)).
$$
Note that this means that $a$ is a $(k-1)$-ary root of $\mathrm{D}p$. If we on the other hand have $\mathrm{D}^k p(a) = 0$ and $\mathrm{D}^{k+1} p(a) \neq 0$, then $\mathrm{D}^kp(x)=(x-a)q(x)$ is integrable and thus
$$\begin{align*}
\mathrm{D}^{k-1} p(x) &= \int \mathrm{D}^k p(x) \mathrm{d}x \\
&= \int (x-a)q(x) \mathrm{d}x \\
&= (x-a)^2 \frac{q(x)}{2} - \int (x-a)^2 \frac{\mathrm{D}q(x)}{2} \mathrm{d}x \\
&= (x-a)^2 \frac{q(x)}{2} - (x-a)^3 \frac{\mathrm{D}q(x)}{3!} + \int (x-a)^3 \frac{\mathrm{D^2}q(x)}{3!} \mathrm{d}x \\
&\vdots\\
&=(x-a)^2 \sum_{k=0}^{\mathrm{deg}(q)} (-1)^k(x-a)^k \frac{\mathrm{D}^kq(x)}{k!} \quad (+C)
\end{align*}
$$
so $\mathrm{D}^{k-1}p$ has a double root at $a$ if $\mathrm{D}^k p$ has a singular one.
So to show that 1 is a triple root of your polynomial just show that the polynomial and it's first few derivatives are $0$.
A: $$\begin{align*}
x^{2n+1}&-(2n+1)x^{n+1}+(2n+1)x^n-1\\
&=x^{2n+1}-1-(2n+1)[ x^{n+1}- x^n]\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+1)-(2n+1)(x-1)x^n\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+x^{n+1}+x^{n-1}+\cdots+1-2nx^n)\\
&=(x-1)[x^n(x^n-1)+x^{n-1}(x^n-x)+\cdots+x^n(x-1)\\
&\qquad\qquad+x^{n-1}(1-x)+x^{n-2}(1-x^2)+\cdots+x(1-x^{n-1})+1-x^n]
\end{align*}$$
Combining corresponding terms from opposite ends of the expression inside the square bracket we have:
$$
=(x-1)[(x^n-1)^2+x(x^{n-1}-1)^2+...+x^{n-1}(x-1)^2]
$$
It is clear that each term inside the square bracket is divisible by $(x-1)^2$ so the whole expression is divisible by $(x-1)^3$
A: Let $f_n(x)=x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$. We will prove the statement by induction on $n$.
For $n=1$ we get $f_1(x)=x^3-3x^2+3x-1=(x-1)^3$ and the statement holds.
Now assume $f_n(x)$ has a factor $(x-1)^3$, then take
\begin{align}
f_{n+1}(x)-xf_n(x)
&=x^{2n+3}-x^{2n+2}-2x^{n+2}+2x^{n+1}-1+x\\
&=x^{2n+2}(x-1)-2x^{n+1}(x-1)+(x-1)\\
&=(x-1)(x^{2n+2}-2x^{n+1}+1)\\
&=(x-1)(x^{n+1}-1)^2\\
&=(x-1)^3\left(\frac{x^{n+1}-1}{x-1}\right)^2\\
&=(x-1)^3\left(1+x+x^2+\dots+x^n\right)^2.
\end{align}
The RHS has a factor $(x-1)^3$, so does $f_n(x)$ (by the induction hypothesis), hence $f_{n+1}(x)$ has a factor $(x-1)^3$.
