Zeros of a polynomial in a simple form Let $m\geq 3$ be a natural number and define
$$f(x)=1-x^m-(1-x)(1+x)^{m-1}.$$
Then, if $m$ is even, the only real roots of $f$ are $x=-1,0,1$ and if $m$ is odd, the roots are $x=0,1$.
My work: for the case where $m$ is even, i proved that
$$f(x)=\sum_{k=0}^{m-2}\left(\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}\right)x^{m-k-1}$$
on one hand. Decomposing it as $f(x)=(x-1)x(x+1)P(x)$, where
\begin{align*}
    P(x)=\sum_{k=0}^{m-4}b_kx^{m-k-4},
\end{align*}
I get an alternative representation
\begin{align*}
    f(x)=\sum_{k=0}^{m-2}(b_k-b_{k-2})x^{m-k-1},
\end{align*}
with $b_{-2}=b_{-1}=b_{m-2}=b_{m-1}=0$. Combining both I get
\begin{align*}
    b_k-b_{k-2}=\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}.
\end{align*}
From here I can clearly see that $b_k\geq 0$, however this does not imply that $P>0$ as intended. I don't know how to proceed, any suggestions please? It is entirely possible that I approached the problem from the wrong perspective...
 A: I was able to prove that $f$ has no more positive roots. Indeed, I needed also to study the sign of this polynomial, so I end up proving a bit more. I used the following reasoning:
\begin{align*}
f(l)=&1-l^m-(1-l)\sum_{k=0}^{m-1}
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k-1}\\
    =&1-l^m-\left[\sum_{k=0}^{m-1}\begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k-1}-\sum_{k=0}^{m-1}\begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k}\right]\\
    =&1-l^m-\left[1+\sum_{k=0}^{m-2}\begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k-1}-\left(\sum_{k=1}^{m-1}\begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k}+l^m\right)\right]\\
    =&\sum_{k=0}^{m-2}\left(\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}\right)l^{m-k-1}\\
    =&\sum_{k=0}^{m-2}\frac{m-2k-2}{k+1}
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}l^{m-k-1}
\end{align*}
We will study the sign of these coefficients. It suffices to study the sign of $a_k:=m-2k-2$ for $k=1, ..., m-2$. If $m$ is even, then $a_k>0$ for $k=0, ..., m/2-2$, $a_{m/2-1}=0$ and $a_k<0$ for $k=m/2, ..., m-2$. On the other hand, if $m$ is odd, then $a_k>0$ for $k=0, ..., (m-3)/2$ and $a_k<0$ for $k=(m-1)/2, ..., m-2$.
The important property is that these coefficients only change sign once, hence by Descartes' rule of signs, $f$ has, at most, one positive root. Since clearly $f(1)=0$ and $f(0)=0$, $f$ does not change sign in the interval $(0,1)$. We will now proceed to prove it is negative in this interval.
Going back to the equality
\begin{align}\label{Equation_expression_polynomial}
    f(l)=\sum_{k=0}^{m-2}\left(\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}\right)l^{m-k-1}.
\end{align}
Now we observe the following. If $m\geq 4$ is even, then $f(-1)=f(0)=f(1)$, hence we can write
\begin{align*}
    f(l)=(l-1)l(l+1)P(l).
\end{align*}
we will prove that $P(l)>0$, which will imply $f(l)\leq 0$ for $0\leq l \leq 1$. Let
\begin{align*}
    P(x)=\sum_{k=0}^{m-4}b_kx^{m-k-4},
\end{align*}
then one can write $f$ in the alternative form
\begin{align*}
    f(x)=\sum_{k=0}^{m-2}(b_k-b_{k-2})x^{m-k-1},
\end{align*}
with $b_{-2}=b_{-1}=b_{m-2}=b_{m-3}=0$. Comparing with \eqref{Equation_expression_polynomial}, we get
\begin{align*}
    b_k-b_{k-2}=\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}.
\end{align*}
Satisfying the symmetry $b_k=b_{m-k-4}$. Indeed, recalling that
\begin{align*}
    c_k:&=\begin{pmatrix}
      m-1\\
      k+1
    \end{pmatrix}-
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}=\frac{m-2k-2}{k+1}
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix}\\
    &=\frac{a_k}{k+1}
    \begin{pmatrix}
      m-1\\
      k
    \end{pmatrix},
\end{align*}
we can use the obvious anti-symmetry $c_k=-c_{m-k-2}$ and an inductive reasoning to imply $b_k=b_{m-k-4}$.
Now we can prove again by inductive reasoning that $b_k>0$, hence $P(l)>0$ for $l\in(0,1)$. This finally implies that $f(l)\leq 0$ for $l\in[0,1]$, as intended.
