The polynomial $x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$ has no real roots. 
Prove that the polynomial $$x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$$ has no real roots.

Here is the solution
Transcribed from this image


  
*For $x \leq 0$ we have obviously $p(x)>0$. Let $x>0$. We transform the polynomial in the same way as a geometric series:
  $$
\begin{aligned}
p(x) &=x^{2 n}-2 x^{2 n-1}+3 x^{2 n-2}-\cdots-2 n x+2 n+1 \\
x p(x) &=x^{2 n+1}-2 x^{2 n}+3 x^{2 n-1}-4 x^{2 n-2}+\cdots+(2 n+1) x.
\end{aligned}
$$
  Adding, we get
  $$
\begin{aligned}
x p(x)+p(x) &=x^{2 n+1}-x^{2 n}+x^{2 n-1}-x^{2 n-2}+\cdots,+x+2 n+1 \\
(1+x) p(x) &=x \cdot \frac{1+x^{2 n+1}}{1+x}+2 n+1.
\end{aligned}
$$
  From here we see that $p(x)>0$ for $x>0$.
  

Can anyone explain how the last line of the solution implies that the polynomial has no real roots?
 A: Since other users already explained the last part of the given proof, I propose another way where there is no need to distinguish between the positive and the negative case, and it does not involve any geometric sum.
By splitting the terms of even degree in a convenient way, the polynomial can be written as
$$(0+1)x^{2n} - 2x^{2n-1} + (1+2)x^{2n-2}- \cdots ((n-1)+n)x^2-2nx + (n + (n+1))$$
that is
$$\left(x^n-x^{n-1}\right)^2+2\left(x^{n-1}-x^{n-2}\right)^2+\dots+n\left(x-1\right)^2+\left(n+1\right)$$
which implies that its minimum value over $\mathbb{R}$ is $n+1>0$ (attained for $x=1$). 
Therefore the polynomial has no real roots.
A: Solution shows that $p(x)>0, \text{ for all } x$, which means that there is no any value of $x$ such that $p(x)=0$ 
A: You've already showed that $p(x)$ can't have any negative roots, and obviously $0$ also isn't a root.
In the last equation, when $x \gt 0$, every term (except possibly $p(x)$) is strictly positive, so dividing through by $1+x$ (which is positive when $x \gt 0$) we see that in this case $p(x)$ also must be strictly positive and $p(x)$ can't have any positive roots either.
We've run out of possibilities for real roots, so $p(x)$ can't have any.
