In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
f(-x) = (-x)³ - 2 (-x)² + 2 (-x) - 6
negative real zeroes = ?
Step 02:
f(-x) = (-x)³ - 2 (-x)² + 2 (-x) - 6
(-x)³ = -x³
(-x)² = x²
(-x) = -x
f(-x) = (-x)³ - 2 (-x)² + 2 (-x) - 6
= -x³ +2x² -2x - 6
no yes no
The answer is:
There are 1 sign changes in f(-x) so there are 1 or 0 possible negative real zeroes.