Taylor approximation is different from the actual value The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$. 
I found that the $3$rd Taylor polynomial with remainder term is \begin{align*}
x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos(2)-4\sin(2)][x+2]^{2}+ \\ 
& [2\cos(2)+  \dfrac{1}{3}\sin(2)][x+2]^{3} + [\dfrac{(x+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
approximating $x=-1.09$ using the Taylor polynomial,
\begin{align*}
(-1.09)^{2}\cos(-1.09) =& 4\cos(2)+[-4\cos(2)+4\sin(2)][(-1.09)+2]+[-\cos(2)-4\sin(2)][(-1.09)+2]^{2}+ \\ 
& [2\cos(2)+  \dfrac{1}{3}\sin(2)][(-1.09)+2]^{3} + [\dfrac{((-1.09)+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]  \\
=& 0.0938985 + [\dfrac{((0.91)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
However, $(-1.09)^{2}\cos(-1.09)=0.549479$ which is a lot of decimals different from Taylor approximation of $0.0938985$.
Is there something that I miss?

This just means that teh third order is not sufficient in particular because $-1.09$ is very far away from $-2$.
To make the story short, what you have is
$$x^2 \cos(x)=\sum_{n=0}^\infty \frac{\left(-n^2+n+4\right) \cos \left(2-\frac{\pi  n}{2}\right)+4 n \sin   \left(2-\frac{\pi  n}{2}\right)}{n!}(x+2)^n$$ So, if you truncate to $O((x+2)^{p+1})$ you will have the following results
$$\left(
\begin{array}{cc}
 2 & 0.4926841712 \\
 3 & 0.0938985194 \\
 4 & 0.6047229336 \\
 5 & 0.5723469909 \\
 6 & 0.5466012534 \\
 7 & 0.5489493571 \\
 8 & 0.5495411050 \\
 9 & 0.5494858556 \\
 10 & 0.5494781119 \\
 11 & 0.5494788050 \\
 12 & 0.5494788702 \\
 13 & 0.5494788647 \\
 14 & 0.5494788644
\end{array}
\right)$$
