2.186. $\frac{p^{3}+4 p^{2}+10 p+12}{p^{3}-p^{2}+2 p+16} \cdot \frac{p^{3}-3 p^{2}+8 p}{p^{2}+2 p+6}$.
## Solution.

Domain of definition: $p \neq-2$.

Factor the numerator of the first fraction:

$$
\begin{aligned}
& p^{3}+4 p^{2}+10 p+12=\left(p^{3}+2 p^{2}\right)+\left(2 p^{2}+4 p\right)+(6 p+12)=p^{2}(p+2)+ \\
& +2 p(p+2)+6(p+2)=(p+2)\left(p^{2}+2 p+6\right)
\end{aligned}
$$

Similarly for the denominator, we find that

$$
\begin{aligned}
& p^{3}-p^{2}+2 p+16=(p+2)\left(p^{2}-3 p+8\right) \text { Then } \\
& \frac{p^{3}+4 p^{2}+10 p+12}{p^{3}-p^{2}+2 p+16} \cdot \frac{p^{3}-3 p^{2}+8 p}{p^{2}+2 p+6}=\frac{(p+2)\left(p^{2}+2 p+6\right)}{(p+2)\left(p^{2}-3 p+8\right)} \times \\
& \times \frac{p\left(p^{2}-3 p+8\right)}{p^{2}+2 p+6}=p
\end{aligned}
$$

Answer: $p$.