Question 74256
<pre>

State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10x-4.


Descartes Rule can help you.

    g(x) = 9x^3 - 7x2 + 10x - 4
           +    -     +     -        Notice the signs
   the signs varies 3 time + -, - +, + - therefore there are 3 variations
   That means that there are 3 possible positive real zeros or 1 positive
   real zeros

Now 
   g(-x) = 9(-x)^3 - 7(-x)^2 + 10(-x) - 4
         = -9x^3 + 7x^2 - 10x - 4
         = -     +      -    -            Notice the signs
   The signs varies 2 times - +, + - therefore there are 2 variations
   that means that there are 2 possible negative zeros


With the given variations the possible zeros are:

  3 positive real zeros, 0 negative real zeros, 0 imaginary
or
  1 positive real zeros, 2 negative real zeros, 0 imaginary

Remember that the highest degree x^3 determines the number of zeros you have
so each total possible zeros are 3

I hope this help