Enter the number of complex zeros for the polynomial function in the box. f(x)=x^3−96x^2+400

F(x) = x³ - 96x² + 400

Apply Descartes' Rule of Signs.
The number of changes in sign is 2, so there are 
(i) either 2 positive real zeros, or
(ii) a pair of conjugate complex zeros.

f(-x) = -x³ - 96x² + 400
The number of changes in sign is 1.
There is one negative real zero.

To determine whether there are 2 real zeros, we should test for zero crossings.
   x          f(x)
----    ----------
  -4     -1200
  -2            8
   0        400
   2           24
   4      -1072

There are 2 zero crossings, therefore there are 2 real zeros and no complex zeros.
The graph of f(x) confirms the analysis.

0

Cbottom left is the correct one.

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