Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$. Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$.
I am thinking about using $Horner's$ $method$ but don't know how.
Let's $P(x) = x^{2n}-nx^{n+1} +nx^{n-1}-1$, we have
$$P'(x) = 2nx^{2n-1}-n(n+1)x^{n}+(n+1)nx^{n-2}$$
$$P''(x) = 2nx^{2n-1}x^{2n-2}-n^2(n+1)x^{n-1}+(n-2)(n-1)nx^{n-3}$$
As $$P(1) = 0$$
$$P'(1) = 0$$
$$P''(1) = 0$$
Then $P(x)$ has root 1 with multiplicity 3
PS: If the OP wants to prove that $P(x)$ has root 1 with multiplicity exactly 3, it suffices to calculate $P'''(x)$ and prove that $P'''(1)  \ne 0$
$$x^{2n}-nx^{n+1} +nx^{n-1}-1$$
$$ x^{2n}-1 +nx^{n-1}-n^{n+1}$$
$$(  x^{2n}-1) + (nx^{n-1}-n^{n+1})$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1) + nx^{n-1}( 1-x^2)$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1) - nx^{n-1}( x+1)(x-1)$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1 - nx^{n-1}( x+1) )$$