1. The problem is to factor the quadratic expression $3x^2 - 16x - 12$ using the middle term break method. 2. The middle term break method involves finding two numbers that multiply to $a \times c$ and add to $b$ in the quadratic $ax^2 + bx + c$. 3. Here, $a = 3$, $b = -16$, and $c = -12$. Calculate $a \times c = 3 \times (-12) = -36$. 4. Find two numbers that multiply to $-36$ and add to $-16$. These numbers are $-18$ and $2$ because $-18 \times 2 = -36$ and $-18 + 2 = -16$. 5. Rewrite the middle term $-16x$ as $-18x + 2x$: $$3x^2 - 18x + 2x - 12$$ 6. Group terms: $$(3x^2 - 18x) + (2x - 12)$$ 7. Factor each group: $$3x(x - 6) + 2(x - 6)$$ 8. Factor out the common binomial factor: $$(3x + 2)(x - 6)$$ 9. Therefore, the factorization of $3x^2 - 16x - 12$ is $$(3x + 2)(x - 6)$$.