fully factor x^3-3x^2+5x-6

To factor the given expression x^3 - 3x^2 + 5x - 6, we can use the Rational Root Theorem and synthetic division to find the roots.

First, we set up the potential roots using the factors of the constant term -6 divided by the factors of the leading coefficient 1. The potential rational roots are ±1, ±2, ±3, and ±6.

By using synthetic division, we find that x = 1 is a root of the polynomial. This means that (x - 1) is a factor of x^3 - 3x^2 + 5x - 6.

Performing synthetic division by dividing x^3 - 3x^2 + 5x - 6 by (x - 1) yields:

1 | 1 -3 5 -6
------------------
1 -2 3 -3

The result of the division gives us x^2 - 2x + 3. To factor this quadratic further, we use the quadratic formula:

x = [2 ± sqrt((-2)^2 - 4*1*3)] / 2
x = [2 ± sqrt(4 - 12)] / 2
x = [2 ± sqrt(-8)] / 2
x = [2 ± 2i√2] / 2
x = 1 ± i√2

Therefore, the fully factored form of x^3 - 3x^2 + 5x - 6 is:

(x - 1)(x - (1 + i√2))(x - (1 - i√2))