Zeroes of 4x3-5x2+6x-3? When x=0, this expression is negative (-3), and when x=1, it is positive (2), so there is a zero between x=0 and x=1.
We can use Newton's Iteration Method to find the zero: xn+1=xn-(4xn3-5xn2+6xn-3)/(12xn2-10xn+6), where the denominator is the derivative of the numerator. The subscript is an iteration number, and to start the process we can use x0=0, so x1=3/6=½. Plug this result into the equation and we get:
x2=11/16, x3=0.6746..., x4=0.67456..., x5=0.674563, which is a stable value. So 0.674563 is a zero.
If we divide by this zero we get the quadratic 4x2-2.3017x+4.4473 (approx) which has only complex zeroes, so the only zero is 0.674563 approx.