The product of the distinct roots of $|x^2 − x − 6| = x + 2$ is

We have: $|x^2 - x - 6| = x + 2$

When we have an equation of the form $|A| = B$, the expression inside the absolute value bars can be either positive or negative, but the result is always $B$.

This gives us two cases:

Case 1: $x^2 - x - 6 = x + 2$ (when the expression is positive)

Case 2: $x^2 - x - 6 = -(x + 2)$ (when the expression is negative)

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$x^2 - x - 6 = x + 2$

$x^2 - x - 6 - x - 2 = 0$

$x^2 - 2x - 8 = 0$

We need two numbers that multiply to $-8$ and add to $-2$. These numbers are $-4$ and $+2$ since $-4 \times 2 = -8$ and $-4 + 2 = -2$.

$(x - 4)(x + 2) = 0$

Therefore: $x = 4$ or $x = -2$

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

$x^2 - x - 6 = -x - 2$

$x^2 - x - 6 + x = -2$

$x^2 - 6 = -2$

$x^2 = 4$

Therefore: $x = 2$ or $x = -2$

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Since we have $|x^2 - x - 6| = x + 2$, the right side must be non-negative.

This means $x + 2 \geq 0$, so $x \geq -2$.

Checking our solutions:

$x = 4$: $4 + 2 = 6 \geq 0$

$x = -2$: $-2 + 2 = 0 \geq 0$

$x = 2$: $2 + 2 = 4 \geq 0$

All solutions are valid!

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From our cases, we found: $x = 4, -2, 2, -2$

Distinct roots means we only count each unique value once.

The distinct roots are: $4, 2, -2$

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Product of distinct roots = $4 \times 2 \times (-2) = -16$

We multiply all unique solutions together, counting each distinct value only once, regardless of how many times it appeared in our cases.