To factor the polynomial \( P(x) = x^5 + 2x^4 + 3x^3 + 3x^2 + 2x + 1 \), we can start by checking for possible rational roots using the Rational Root Theorem. The Rational Root Theorem states that any possible rational root of a polynomial with integer coefficients is a factor of the constant term divided by a factor of the leading coefficient.

For the polynomial \( P(x) = x^5 + 2x^4 + 3x^3 + 3x^2 + 2x + 1 \):
- The constant term is \( 1 \).
- The leading coefficient is \( 1 \).

The possible rational roots are therefore \( \pm 1 \).

Let's test these possible roots:

1. Testing \( x = 1 \):
\[
P(1) = 1^5 + 2 \cdot 1^4 + 3 \cdot 1^3 + 3 \cdot 1^2 + 2 \cdot 1 + 1 = 1 + 2 + 3 + 3 + 2 + 1 = 12 \neq 0
\]
So, \( x = 1 \) is not a root.

2. Testing \( x = -1 \):
\[
P(-1) = (-1)^5 + 2 \cdot (-1)^4 + 3 \cdot (-1)^3 + 3 \cdot (-1)^2 + 2 \cdot (-1) + 1 = -1 + 2 - 3 + 3 - 2 + 1 = 0
\]
So, \( x = -1 \) is a root.

Since \( x = -1 \) is a root, we can factor \( P(x) \) as \( (x + 1) \) times a polynomial of degree 4. We can perform polynomial division to find the quotient.

Performing synthetic division of \( P(x) \) by \( x + 1 \):

\[
\begin{array}{r|rrrrrr}
-1 & 1 & 2 & 3 & 3 & 2 & 1 \\
& & -1 & -1 & -2 & -1 & -1 \\
\hline
& 1 & 1 & 2 & 1 & 1 & 0 \\
\end{array}
\]

The quotient is \( x^4 + x^3 + 2x^2 + x + 1 \), so we have:
\[
P(x) = (x + 1)(x^4 + x^3 + 2x^2 + x + 1)
\]

Next, we need to factor \( x^4 + x^3 + 2x^2 + x + 1 \). We can check for rational roots again, but this polynomial does not have any obvious rational roots. We can use other methods such as factoring by grouping or applying more advanced techniques.

However, it turns out that \( x^4 + x^3 + 2x^2 + x + 1 \) does not factor nicely over the rationals. Therefore, the factorization of \( P(x) \) over the rationals is:

\[
P(x) = (x + 1)(x^4 + x^3 + 2x^2 + x + 1)
\]

This is the simplest factorization we can achieve using elementary methods. Further factorization would require more advanced techniques or numerical methods.