Rules for Landau Symbols

Question

Let $x \in \mathbb{R}$. Then I have the following \begin{align} e^x (12 - 6x + x^2) &= \left((1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \mathcal{O}(x^5)\right)(12 - 6x + x^2) \\ &= 12 - 6x + 6x^2 + \mathcal{O}(x^5). \end{align}

However, I don't get the last equation. Isn't $ x \mathcal{O}(x^n) = \mathcal{O}(x^{n+1})$?

Answer 1

Note that the coefficient of $x^5$ in $$ e^x (12 - 6x + x^2) $$

$$= \left(1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 +\frac{1}{120}x^5+...\right)(12 - 6x + x^2) \\ $$

$$= 12 - 6x + 6x^2 + \frac {1}{60}x^5+...$$

$$= 12 - 6x + 6x^2 + \mathcal{O}(x^5)$$