Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x^3-1)(x^2-x+1)\\
&=(x-1)(x^2+x+1)(x^2-x+1).
\end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$? 

The derivative has a single real root, namely $1$. The other two factors never vanish on $\mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-\infty$ and $\infty$ are both $\infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}=\frac{30-20+15-12+10}{60}=\frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
