SOLUTION: State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10-4 Thank you s much for all of your help!!!!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10-4 Thank you s much for all of your help!!!!      Log On


   

Question 346076: State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10-4
Thank you s much for all of your help!!!!
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you mean g%28x%29=9x%5E3-7x%5E2%2B10x-4
.
.
.
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C9x%5E3-7x%5E2%2B10x-4%29
.
.
.
From the graph, there is one positive real zero, which means the remaining two roots are imaginary.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10x-4
-----------------
# of sign changes in g(x) is 3 ; so 3 or 1 positive Real zeros.
----
# of sign changes in g(-x) is 0 ; so no negative Real zeros.
-----
Fact: There is one positive around x = 0.4606
There are no negative so there are 2 imaginary.
===============================
Cheers,
Stan H.