Task A-1.1. (8 points)

Simplify the fraction $\frac{a^{4}-2 a^{3}-2 a^{2}+2 a+1}{(a+1)(a+2)}$.
## Solution.

$$
\begin{aligned}
\frac{a^{4}-2 a^{3}-2 a^{2}+2 a+1}{(a+1)(a+2)} & =\frac{a^{4}-2 a^{2}+1-2 a^{3}+2 a}{(a+1)(a+2)}=\frac{\left(a^{2}-1\right)^{2}-2 a\left(a^{2}-1\right)}{(a+1)(a+2)} \\
& =\frac{\left(a^{2}-1\right)\left(a^{2}-1-2 a\right)}{(a+1)(a+2)}=\frac{(a-1)(a+1)\left(a^{2}-1-2 a\right)}{(a+1)(a+2)} \\
& =\frac{(a-1)\left(a^{2}-2 a-1\right)}{a+2} \text { or } \frac{a^{3}-3 a^{2}+a+1}{a+2}
\end{aligned}
$$

For 2 points: the factor $(a+1)$ was factored out in the numerator, but there is an error in the second parenthesis.

For 8 points: the correct result and procedure.