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?

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> 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?      Log On


   

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) About Me  (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.
graph%28300%2C200%2C-10%2C10%2C-10%2C10%2C7x%5E3-3x%5E2%2B8x-1%29