Question 977610
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(300,200,-10,10,-10,10,7x^3-3x^2+8x-1)}}}