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?

Expand just the cosine series around $x=-2$ and you will have:
$$x^2cos x = x^2\left(cos(2) + sin(2)[x+2] - \frac{cos(2)[x+2]^2}{2} - \frac{sin(2)[x+2]^3}{6}\right )$$
$$\\$$
(Expanding the $x^2$ in a Taylor polynomial and multiplying everything by brute force also works, but the Taylor series for $x^2$ will just reduce back to $x^2$ anyway, so it is not necessary)
$$\\$$
Evaluated at $x=-1.09$: 
$$ (-1.09)^2cos(-1.09) \approx 0.549479$$
$$\\$$
$$(-1.09)^2\left(cos(2) + sin(2)[0.91] - \frac{cos(2)[0.91]^2}{2} - \frac{sin(2)[0.91]^3}{6}\right )\approx 0.557713$$
