SOLUTION: using descartes rule of signs how positive,negative, and real zeros the polynomial P(x)=7x^3-3x^2+8x-1 number of positive zeros number of negative zeros? number of real zeros?

Question 977610: using descartes rule of signs how positive,negative, and real zeros the polynomial P(x)=7x^3-3x^2+8x-1
number of positive zeros
number of negative zeros?
number of real zeros?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
7x^3-3x^2+8x-1
There are 3 sign changes, so there can be 3 or 1 positive zeros.
make x negative
7(-x)^3-3(-x^2)-8(-x) -1
-7x^3-3x^2+8x-1. There are 2 sign changes, so there can be 2 or 0 negative zeros.
There is one real zero, and it is positive. The other two roots are complex.

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