Factorize This

$\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}$

$\Rightarrow x^2 - 19 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)$

$\begin{cases} x=1\quad \quad \Rightarrow \quad 18=-6A\quad \, \, \, \Rightarrow \quad A=3 \\ x=-2\quad \, \Rightarrow \quad -15=15B\quad \Rightarrow \quad B=-1 \\ x=3\quad \quad \Rightarrow \quad -10=10C \quad \Rightarrow \quad C=-1 \end{cases}$

$\Rightarrow ABC = -1 \times 3 \times -1 = \large \color{#20A900}{\boxed{3}}$

Relevant wiki: Partial Fractions

$\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}$

$\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)}{(x-1)(x+2)(x-3)}$

$\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x^2 - x -6) + B(x^2 - 4x + 3) + C(x^2 + x - 2)}{(x-1)(x+2)(x-3)}$

$\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{(A+B+C)x^2 + (C-A-4B)x + (3B-6A-2C)}{(x-1)(x+2)(x-3)}$

Then we can set equations for the coefficients of the numerator:

$\begin{cases}A+B+C=1\\C-A-4B=0\\3B-6A-2C= -19\end{cases}$

Adding the first two equations, we will get $2C-3B = 1$ , and by adding this to the last equation, we will come up with: $-6A -1 = -19$ . Thus, $A =3$ .

Then, $C = -B-2$ .

$-B-2-3-4B = -5B-5 = 0$ . Thus, $B= -1$ .

Hence, $C=-1$ .

Therefore, $A\times B\times C = 3\times -1\times -1 = \boxed{3}$ .

$\begin{aligned} \frac{x^2-19}{x^3-2x^2-5x+6} & = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \\ \frac{x^2-19}{(x-1)(x+2)(x-3)} & = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }(x-1)(x+2)(x-3)} \\ x^2 - 19 & = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) \\ \color{#3D99F6}{\text{Let }x=1:} \quad 1 - 19 & = A(3)(-2) + B(0) + C(0) \\ \implies \color{#3D99F6}{A} & = \color{#3D99F6}{3} \\ \color{#3D99F6}{\text{Let }x=-2:} \quad 4-19 & = B(-3)(-5) \\ \implies \color{#3D99F6}{B} & = \color{#3D99F6}{-1} \\ \color{#3D99F6}{\text{Let }x=3:} \quad 9-19 & = C(2)(5) \\ \implies \color{#3D99F6}{C} & = \color{#3D99F6}{-1} \end{aligned}$

$\implies \color{#3D99F6}{A} \times \color{#3D99F6}{B} \times \color{#3D99F6}{C} = (\color{#3D99F6}{3})(\color{#3D99F6}{-1})(\color{#3D99F6}{-1}) = \boxed{3}$

Using cover up method from wiki we get the following.
$A=\dfrac{x^2-19}{(x+2)(x-3)}|_{x=1}=\dfrac{1^2-19}{(1+2)(1-3)}=3.\\ B=\dfrac{x^2-19}{(x-1)(x-3)}|_{x=-2}=\dfrac{(-2)^2-19}{(-2 - 1)(-2 - 3)}=-1.\\ C=\dfrac{x^2-19}{(x-1)(x+2)}|_{x=3}=\dfrac{3^2-19}{(3 - 1)(3+2)}= - 1.\\ \therefore\ A*B*C=3$