Question 166495
State the number of positive real zeros, negative real zeros, and imaginary zeros for 
g(x)=9x^3-7x^2+10x-4.
g(-x) = -9x^2 - 7x^2 - 10x -4
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Using DesCartes Rule of Signs:
# of changes of sign in g(x) = 3; so 1 or 3 positive Real roots
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# of changes of sign in g(-x) = 0; so 0 negative Real Roots
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# of imaginary zeroes is 0 or two  because g(x) is a cubic
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Comment: graphing g(x) is find it has one positive Real Root.
So it must have 2 imaginary roots.
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Cheers,
Stan H.