factor completely : x^3-2x^2+6x-12?

[05]
x^3-2x^2+6x-12
=x^2(x-2)+6(x-2)
=(x-2)(x^2+6)

f(2) = 8 - 8 + 12 - 12 = 0
Thus x - 2 is a factor of f(x)
To find other factors use synthetic division:-
2|1____-2____6____-12
_|_____2____0_____12
_|1____0____6_____0

(x - 2)(x² + 6)

(x^2+6)(x-2)

x^3 - 2x^2 + 6x - 12
= (x^2 + 6)(x - 2)

(x^2 + 6 ) ( x - 2 )

2 is a root through guessing and checking
Dividing the equation by (x-2) gives:
(x^2 + 6), which can't be factored.

(x-2)(x^2 + 6)

By synthetic division:

2 _| 1 -2 6 -12
___2_0__12_
1 0 6 0

The factors are (x-2)(x^2 + 6).

x^3-2x^2+6x-12
(x^2 + 6 ) ( x - 2 )

x³ - 2x² + 6x - 12
Group:
= (x³ - 2x²) + (6x - 12)
Factor each term:
= x²(x - 2) + 6(x - 2)
Take out an (x-2):
= (x - 2)(x² + 6)