completely: $$ 6 x^{2}+12 x-48 $$

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To **completely factor** the quadratic expression $$ 6x^{2} + 12x - 48 $$, follow these steps:

1. **Factor Out the Greatest Common Factor (GCF):**
   
   First, identify the GCF of the coefficients $$6$$, $$12$$, and $$-48$$, which is $$6$$.

$$
   6x^{2} + 12x - 48 = 6(x^{2} + 2x - 8)
   $$

2. **Factor the Quadratic Expression Inside the Parentheses:**

Now, factor $$x^{2} + 2x - 8$$. We're looking for two numbers that multiply to $$-8$$ (the constant term) and add up to $$2$$ (the coefficient of $$x$$).

However, since $$-8$$ doesn't have integer factors that add up to $$2$$, the quadratic doesn't factor neatly over the integers. Instead, we'll use the method of factoring by grouping using the original expression:

$$
   6x^{2} + 12x - 48 = (x + 4)(6x - 12)
   $$

3. **Factor Further if Possible:**

Notice that $$6x - 12$$ has a common factor of $$6$$:

$$
   (x + 4)(6x - 12) = 6(x + 4)(x - 2)
   $$

Thus, the expression is fully factored.

**Final Factored Form:**

$$
6(x + 4)(x - 2)
$$
The expression $$6x^{2} + 12x - 48$$ factors to $$6(x + 4)(x - 2)$$.

completely: \( 6 x^{2}+12 x-48 \)

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