What are the factors for the cubic equation: x^3 -8x^2 +19x -12?

Answer 1

others have mentioned the values and let me explain the method
f(x) changes sign 3 times so 1 or 3 positive root
f(-x) = -x^3-8x^2 -19x -12 no change of sign so 0 negative root

if it has a real 0 then it has to be factor of -12 that is 1/-1/2/-2/3/-3/6/-6/12/-12/4/-4
we need to test only positive roots 1 2 3 6 12 or 4
trying x = 1 we get 1-8+19 - 12 =0

so (x-1) is a factor
dy division
x^3-8x^2+19x-12
= x^3-x^2-7x^2+7x + 12 x - 12
= x^2(x-1) - 7x(x-1) + 12(x-1)
= (x-1)(x^2-7x+12)
quadratic we have to factor 12 so that sum is -7 we get -4 - 3

so result = (x-1)(x-3)(x-4)

Answer 2

x^3 -8x^2 +19x -12 can be factored into (x-1)(x-3)(x-4).

Answer 3

(x-1)(x-3)(x-4) = x^3 -8x^2 +19x -12

Answer 4

(x-4)(x-3)(x-1)
you can get this easily if you can find a zero first and then use synthetic division.