at most, it can have 4, and at least it can have 0
anwyay, we can factor to solve for where it equals 0
f(x)=0 is where there is an x intercept
0=x^4-x^3+x^2-x
factor out the x
0=x(x^3-x^2+x-1)
group and factor
0=x((x^3-x^2)+(x-1))
0=x(x^2(x-1)+1(x-1))
0=x(x-1)(x^2+1)
set each to zero
0=x
0=x-1
1=x
0=x^2+1
-1=x^2
false
the x intercepts are at x=0 and x=1 or the points (0,0) and (1,0)
