How many real numbers are roots of the polynomial \[x^9 - 37x^8 - 2x^7 + 74x^6 + x^4 - 37x^3 - 2x^2 + 74x?\]
1. Given the polynomial:
   \[
   P(x) = x^9 - 37x^8 - 2x^7 + 74x^6 + x^4 - 37x^3 - 2x^2 + 74x
   \]
   We need to determine the number of real roots.

2. The polynomial can be factored as:
   \[
   P(x) = (x^4 - 37x^3 - 2x^2 + 74x)(x^5 + 1)
   \]

3. Further factorization of \(x^4 - 37x^3 - 2x^2 + 74x\):
   \[
   x^4 - 37x^3 - 2x^2 + 74x = x(x^3 - 37x^2 - 2x + 74)
   \]

4. The polynomial \(x^3 - 2x\) can be factored as:
   \[
   x^3 - 2x = x(x^2 - 2) = x(x - \sqrt{2})(x + \sqrt{2})
   \]

5. Therefore, the factorization of the original polynomial becomes:
   \[
   P(x) = x(x - \sqrt{2})(x + \sqrt{2})(x - 37)(x^5 + 1)
   \]

6. The roots of \(x(x - \sqrt{2})(x + \sqrt{2})(x - 37)\) are:
   \[
   x = 0, \pm \sqrt{2}, 37
   \]
   These are four distinct real roots.

7. The polynomial \(x^5 + 1\) can be factored using the sum of cubes:
   \[
   x^5 + 1 = (x + 1)(x^4 - x^3 + x^2 - x + 1)
   \]
   The factor \(x + 1\) has a real root:
   \[
   x = -1
   \]
   The factor \(x^4 - x^3 + x^2 - x + 1\) has no real roots because its discriminant is negative, indicating no real solutions.

8. Combining all the real roots, we have:
   \[
   x = 0, \pm \sqrt{2}, 37, -1
   \]

9. Therefore, the total number of real roots is:
   \[
   \boxed{5}
   \]