Question 105829
{{{f(x)=32x^3-52x^2+17x+3}}}
I'm not sure of the questions (discarded?). 
Looking at the graph, there are three zeros. 
One of them is at x=1. 
{{{f(1)=32(1)^3-52(1)^2+17(1)+3=32-52+17+3=0}}}
Another between -1 and 0. The third between 0 and 1. 
Hopefully this will help. 
Please re-post your question. 
{{{ graph( 300, 300, -1, 2, -5, 5, 32x^3-52x^2+17x+3)}}}