Taylor polynomial of degree 2 of $e^{x^2+x}$ I want to find the Taylor polynomial of degree 2 of $e^{x^2+x}$ and this is what the answer should be:
$$e^{x^2+x} = e^{x^2}e^{x} = (1 + x^2 + O(x^4)) (1 + x + \cfrac{x^2}{2} + O(x^3)) = 1 + x + \cfrac{3x^2}{2} + O(x^3)$$
The thing is that I don't understand what happens after the last equals sign. If anyone would care to explain that would be much appreciated.

You multiply the two factors with the usual rules for polynomials (distributivity/associativity), with the additional rules that $O(x^k)\cdot \alpha x^\ell = O(x^{k+\ell})$ and $O(x^k)\cdot O(x^\ell) = O(x^{k+\ell})$:
$$
\begin{align}
(1 + x^2 + O(x^4)) (1 + x + \frac{x^2}{2} + O(x^3)) 
= 1 + x + \frac{x^2}{2} + O(x^3) + x^2 \cdots
\end{align}
$$
except that you stop whenever you get a power bigger than what you get in any of the $O(\cdot)$ so far. This is why here I stopped the expansion and didn't include any $x^3$, $x^4$, or $x^5$: there is no point, as I have a $O(x^3)$ already which "swallows" all these terms. You can go further in the expansion, but in the end any $x^k$ term for $k\geq 3$ will disappear, since they are all hidden in the $O(x^3)$.
Then, you gather the terms:
$$
\begin{align}
1 + x + \frac{x^2}{2} + O(x^3) + x^2= 1+x +\frac{3}{2}x^2 + O(x^3)
\end{align}
$$
