Full question:
Show that if $n$ is even there exists a polynomial of degree $n$ that has no roots in $\mathbb{R}$.
$n=2k$ and $p(x)=x^n-x^{n-1}+2$
Would this polynomial satisfy the equation, I can't think of any real roots?
Your polynomial indeed works, but for the case $x>0$ you have to do some more work, because it's not obvious that $x^n-x^{n-1}+2>0$.
However, $x^n-x^{n-1}+2=x^{n-1}(x-1)+2$ is certainly positive for $x\ge1$ and $x\le0$ (in the latter case because $x^n\ge0$ and $-x^{n-1}\ge0$).
If $0<x<1$, we have $0<x^n<x^{n-1}<1$, so also $0<x^{n-1}-x^n<1$, hence $-1<x^n-x^{n-1}<0$ and $x^n-x^{n-1}+2>1$.
Considering $x^n+1$ is much easier.
Your polynomial does satisfy the problem (but is not the easiest choice).
Let $f_n(x) = x^n + x^{n-1} + 2$. Then $f_n'(x) = x^{n-2}(nx + 1-n)$. Since, for even $n$, $\lim_{x \rightarrow \pm \infty} f_n(x) = \infty$, these $f_n$ each have a global minimum at one of their local minima. By the first derivative, candidates are $x = 0$ and $x = \frac{n-1}{n}$. Since $f_n'(-1) < 0$, $f_n'(1/4) < 0$, and $f_n'(2) > 0$, only $\frac{n-1}{n}$ can be a minimum. Then $$ f_n\left(\frac{n-1}{n}\right) = \frac{2n - 2 - \left(1 - \frac{1}{n}\right)^n}{n-1} $$ The part $\left(1 - \frac{1}{n}\right)^n$ is a standard way to get $\mathrm{e}^{-1}$ (by taking the limit as $n \rightarrow \infty$) and takes values in $[0,1]$. (We can get tighter bounds, but there is no need.) Since $n \geq 2$, the numerator is at least $4-2-1 > 0$ and the denominator is also positive, so the global minimum is positive. Therefore, $f_n(x)$ is positive for even $n$ and all $x \in \mathbb{R}$.
Contrast with: $x^{2k} = (x^2)^k$ is always nonnegative, so $x^{2k}+1$ is always positive.
$x^2+1$.
For your polynomial, I think that it works, but you are complicating it with $x%{n-1}$ since this can be negative.
