Factor the expression completely $$ 3 s^{2}-3 s-18 $$

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To factor the expression $$ 3s^{2} - 3s - 18 $$ completely, follow these steps:

1. **Factor Out the Greatest Common Factor (GCF):**
   
   All the coefficients in the expression are divisible by 3. So, factor out 3:
   
   $$
   3s^{2} - 3s - 18 = 3(s^{2} - s - 6)
   $$

2. **Factor the Quadratic Expression:**
   
   Now, focus on factoring the quadratic expression inside the parentheses: $$ s^{2} - s - 6 $$. We need to find two numbers that multiply to $$-6$$ (the constant term) and add up to $$-1$$ (the coefficient of the middle term).
   
   The numbers $$-3$$ and $$2$$ satisfy these conditions because:
   
   $$
   -3 \times 2 = -6 \quad \text{and} \quad -3 + 2 = -1
   $$
   
   Thus, the quadratic can be factored as:
   
   $$
   s^{2} - s - 6 = (s - 3)(s + 2)
   $$

3. **Combine the Factors:**
   
   Incorporate the GCF back into the expression:
   
   $$
   3(s^{2} - s - 6) = 3(s - 3)(s + 2)
   $$

**Final Factored Form:**

$$
3(s - 3)(s + 2)
$$
The expression $$ 3s^{2} - 3s - 18 $$ factors to $$ 3(s - 3)(s + 2) $$.

Factor the expression completely \( 3 s^{2}-3 s-18 \)

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