The number of real roots of `x^(8) - x^(5) + x^(2) - x + 1 = 0` is

Explanation

To determine the number of real roots of the equation \( x^{8} - x^{5} + x^{2} - x + 1 = 0 \), we can use Descartes' Rule of Signs. This rule helps us find the possible number of positive and negative real roots by analyzing the sign changes in the polynomial.

### Step-by-Step Solution:

1. **Identify the function**: Let \( f(x) = x^{8} - x^{5} + x^{2} - x + 1 \).

2. **Check for positive roots**: We will evaluate \( f(x) \) for positive values of \( x \). We need to count the number of sign changes in the sequence of coefficients of \( f(x) \).

- The coefficients of \( f(x) \) are: \( 1, 0, 0, -1, 1 \).
- The signs of these coefficients are: \( +, 0, 0, -, + \).
- We observe the following sign changes:
- From \( + \) to \( - \) (1 change)
- From \( - \) to \( + \) (1 change)
- Total sign changes for positive roots = 2.

According to Descartes' Rule of Signs, the number of positive real roots can be 2 or 0.

3. **Check for negative roots**: Now we will evaluate \( f(-x) \) to check for negative roots.

- Calculate \( f(-x) \):
\[ f(-x) = (-x)^{8} - (-x)^{5} + (-x)^{2} - (-x) + 1 = x^{8} + x^{5} + x^{2} + x + 1 \]
- The coefficients of \( f(-x) \) are: \( 1, 1, 1, 1, 1 \).
- The signs of these coefficients are all positive: \( +, +, +, +, + \).
- There are no sign changes.

According to Descartes' Rule of Signs, this means there are 0 negative real roots.

4. **Conclusion**:
- From the analysis, we found that there can be either 2 or 0 positive real roots and 0 negative real roots.
- Therefore, the total number of real roots is at most 2.

### Final Answer:
The number of real roots of the equation \( x^{8} - x^{5} + x^{2} - x + 1 = 0 \) is at most 2.