Problem 1052

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:
$$\boxed{5}$$