Given $a, \theta \in \mathbb R, m \in \mathbb N$, and $P(x) = x^{2m}- 2|a|^mx^m \cos \theta +a^{2m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.
1. Given the polynomial \( P(x) = x^{2m} - 2|a|^m x^m \cos \theta + a^{2m} \), we need to factorize it as a product of \( m \) real quadratic polynomials.

2. First, observe that \( P(x) \) can be rewritten in terms of \( y = x^m \):
   \[
   P(x) = (x^m)^2 - 2|a|^m x^m \cos \theta + a^{2m}
   \]
   Let \( y = x^m \). Then the polynomial becomes:
   \[
   P(y) = y^2 - 2|a|^m y \cos \theta + a^{2m}
   \]

3. This is a quadratic equation in \( y \). Solving for \( y \), we use the quadratic formula:
   \[
   y = \frac{2|a|^m \cos \theta \pm \sqrt{(2|a|^m \cos \theta)^2 - 4 \cdot 1 \cdot a^{2m}}}{2 \cdot 1}
   \]
   Simplifying the discriminant:
   \[
   y = \frac{2|a|^m \cos \theta \pm \sqrt{4|a|^{2m} \cos^2 \theta - 4a^{2m}}}{2}
   \]
   \[
   y = \frac{2|a|^m \cos \theta \pm 2|a|^m \sqrt{\cos^2 \theta - 1}}{2}
   \]
   \[
   y = |a|^m \cos \theta \pm |a|^m \sqrt{\cos^2 \theta - 1}
   \]
   Since \(\cos^2 \theta - 1 = -\sin^2 \theta\), we have:
   \[
   y = |a|^m \cos \theta \pm i |a|^m \sin \theta
   \]
   Therefore, the roots are:
   \[
   y_1 = |a|^m (\cos \theta + i \sin \theta)
   \]
   \[
   y_2 = |a|^m (\cos \theta - i \sin \theta)
   \]

4. Now, we need to find the \( m \)-th roots of these complex numbers. Let \( t_1, t_2, \ldots, t_m \) be the \( m \)-th roots of \( y_1 \) and \( s_1, s_2, \ldots, s_m \) be the \( m \)-th roots of \( y_2 \). These roots form regular \( m \)-polygons in the complex plane centered at the origin with radius \( |a| \).

5. Since \( y_1 \) and \( y_2 \) are complex conjugates, their \( m \)-th roots will also be complex conjugates. We can pair these roots as \( (t_1, s_1), (t_2, s_2), \ldots, (t_m, s_m) \).

6. Each pair \( (t_i, s_i) \) forms a quadratic polynomial:
   \[
   f_i(x) = (x - t_i)(x - s_i)
   \]
   Since \( t_i \) and \( s_i \) are complex conjugates, \( f_i(x) \) will be a real quadratic polynomial.

7. Therefore, the factorization of \( P(x) \) is:
   \[
   P(x) = f_1(x) f_2(x) \cdots f_m(x)
   \]
   where \( f_i(x) = (x - t_i)(x - s_i) \) for \( i = 1, 2, \ldots, m \).

The final answer is \( \boxed{ P(x) = f_1(x) f_2(x) \cdots f_m(x) } \) where \( f_i(x) = (x - t_i)(x - s_i) \) for \( i = 1, 2, \ldots, m \).