To find the equation of a straight line with a gradient (slope) of \(-2\) that passes through the point \((4, 1)\), we use the point-gradient form of the line equation:

\[
y - y_1 = m(x - x_1)
\]

where:
- \(m\) is the gradient,
- \((x_1, y_1)\) is the point the line passes through.

For the given line:
- \(m = -2\),
- \((x_1, y_1) = (4, 1)\).

Substitute these values into the point-gradient form:

\[
y - 1 = -2(x - 4)
\]

Expand and simplify:

\[
y - 1 = -2x + 8
\]

\[
y = -2x + 8 + 1
\]

\[
y = -2x + 9
\]

To put this in the standard form \(Ax + By + C = 0\), rearrange:

\[
2x + y - 9 = 0
\]

Conclusion:
The correct equation of the line is C. \(y + 2x - 9 = 0\).