1. 1) Solution. First method. \(a^{4}+2 a^{3}+1=\left(a^{4}+a^{3}\right)+\) \(+\left(a^{3}+1\right)=a^{3}(a+1)+(a+1)\left(a^{2}-a+1\right)=(a+1) \times\) \(\times\left(a^{3}+a^{2}-a+1\right)\).

Second method. \(a^{4}+2 a^{3}+1=a^{4}+a^{3}+\)

\[
\begin{gathered}
+a^{3}+a^{2}- \\
-a^{2}-a+ \\
+a+1= \\
=a^{3}(a+1)+a^{2}(a+1)-a(a+1)+(a+1)= \\
=(a+1)\left(a^{3}+a^{2}-a+1\right) .
\end{gathered}
\]

Third method. \(a^{4}+2 a^{3}+1=\left(a^{4}+2 a^{3}+a^{2}\right)-a^{2}+1=\) \(=a^{2}(a+1)^{2}-(a+1)(a-1)=(a+1)\left(a^{3}+a^{2}-a+1\right)\).

Other methods of solving the problem are possible, but they are usually either cumbersome or require additional knowledge. For example, one could try to divide the given polynomial by \(a+1\).
2) \((m-1)\left(m^{2}+m+3\right)\)
3) \((a-1)(a+1)\left(2 a^{2}-1\right)\).
1. Factorize the polynomials:
1) \(a^{4}+2 a^{3}+1\)
2) \(m^{3}+2 m-3\);
3) \(2 a^{4}-a^{2}-1\).