Given that \( x = 1 \) is a root of the polynomial equation \( x^3 - 2x^2 - 5x + 6 = 0 \), we can find the other roots by performing polynomial division to factorize the cubic polynomial.

1. Divide \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \):

Using synthetic division:

\[
\begin{array}{r|rrrr}
1 & 1 & -2 & -5 & 6 \\
& & 1 & -1 & -6 \\
\hline
& 1 & -1 & -6 & 0 \\
\end{array}
\]

The quotient is \( x^2 - x - 6 \), and the remainder is 0.

2. Factorize \( x^2 - x - 6 \):

We need to factorize \( x^2 - x - 6 \):
\[
x^2 - x - 6 = (x - 3)(x + 2)
\]

3. Combine the factors:

The factorization of the original polynomial is:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)
\]

Therefore, the other roots are \( x = 3 \) and \( x = -2 \).

The correct answer is:

C. 3 and -2