Product of Distinct Roots of Absolute Quadratic

Note $|x^2 - x - 6| = |(x-3)(x+2)| = x + 2$. Case 1: $x + 2 < 0 \implies x < -2$: No real solution since LHS $\ge 0$. Case 2: $x = -2$: LHS $= 0 =$ RHS, so $x = -2$ is a solution. Case 3: $x > -2$: $x+2 > 0$, so we divide both sides by $x+2$. $|x - 3| = 1 \implies x - 3 = 1$ or $x - 3 = -1 \implies x = 4$ or $x = 2$. Both $4$ and $2$ are $> -2$. Distinct roots are $-2, 2, 4$. Product $= (-2) \times 2 \times 4 = -16$.