You can put this solution on YOUR website! is a polynomial of third degree (odd)
in real coefficients.
Hence = 0 must have one real
or three real roots.
When it has only one real root, the other two are complex conjugate
number.
Hence, the possible number of imaginary zeros is 2 or 0.
[In fact, the complex roots appear always in pairs]
In fact, f(0) = -3 < 0 and f(1) = 3 > 0,so there is a zero of f
between 0 and 1. Also, since f'(x) = 6x^2 -8x + 8 = 2(3x^2 -4x + 4) =0
has no real roots, so f has only one real zero and two complex zeros.
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