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 

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