What are the zeros of the polynomial function: f(x) = x3 + x2 – 6x ?

The solution to the problem is as follows:

　 
x³ + x² − 6x = 0 
x (x² + x − 6) = 0 
x (x − 2) (x + 3) = 0 
x = 0, 2, −3 

f(x) = 2x³ + 5x² + 6x − 4 
f(−x) = −2x³ + 5x² − 6x − 4 

Let's look at signs only and how many sign changes there are: 
f(x) = + + + − 
1 sign change > 1 positive root 
f(−x) = − + − − 
2 sign changes > 2 or 0 negative roots 

Min. number of real roots = 1 > max. number of complex roots = 2 
Max. number of real roots = 3 > min. number of complex roots = 0

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The answer would be c