$\displaystyle x^3-3x^2-4x+12 = 0$

$\displaystyle [(x-3)x-4]x+12 = 0

$

$\displaystyle f(-2) = 0$

** ∴ **$\displaystyle x^3-3x^2-4x+12$ ** ≡** $\displaystyle (x+2)(ax^2+bx+c)$

$\displaystyle x^3-3x^2-4x+12$** ≡ **$\displaystyle ax^3+bx^2+cx+2ax^2+2bx+2c$

$\displaystyle x^3-3x^2-4x+12$** ≡ **$\displaystyle ax^3+bx^2+2ax^2+cx+2bx+2c$

$\displaystyle x^3-3x^2-4x+12$ ** ≡ **$\displaystyle ax^3+(b+2a)x^2+(c+2b)x+2c$

$\displaystyle a =1$; $\displaystyle b+2a = -3$, ∴ $\displaystyle b = -5$; $\displaystyle c+2b -4$, ∴ $\displaystyle c = 6$.

** ∴ **$\displaystyle x^3-3x^2-4x+12$ ** ≡ ** $\displaystyle (x+2)(x^2-5x+6)$

$\displaystyle x^2-5x-6 = 0$

$\displaystyle (x-5)x-6 = 0$

$\displaystyle f(2) = 0$

∴ $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(ax+b)$

$\displaystyle x^2-5x-4 $ ≡ $\displaystyle ax^2+bx-2ax-2b$

$\displaystyle x^2-5x-4$ ≡ $\displaystyle (a)x^2+(b-2a)x-2b$

$\displaystyle a = 1$; $\displaystyle b-2a$ = $\displaystyle -5$, ∴ $\displaystyle b = -3$.

∴ $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(x-3)$

∴ $\displaystyle x^3-3x^2-4x+12$ ≡ $\displaystyle (x+2)(x-2)(x-3)$