To factor the quadratic expression $3x^2 - 4x - 4$, let's follow these steps:

1. Identify the coefficients:

   • $a = 3$
   • $b = -4$
   • $c = -4$

2. Multiply $a$ and $c$:

   • $3 \times (-4) = -12$

3. Find two numbers that multiply to -12 and add to -4:

   • The numbers that satisfy this are $-6$ and $2$, because $-6 \times 2 = -12$ and $-6 + 2 = -4$.

4. Rewrite the middle term using these numbers: $3x^2 - 6x + 2x - 4$

5. Factor by grouping:

   • Group terms: $(3x^2 - 6x) + (2x - 4)$
   • Factor out the common factors in each group: $3x(x - 2) + 2(x - 2)$

6. Factor out the common binomial:

   • $(3x + 2)(x - 2)$

So, the fully factored form of the expression $3x^2 - 4x - 4$ is: $(3x + 2)(x - 2)$

Would you like further details or explanations on this process?

Here are some additional questions related to this topic:

1. How would you factor a quadratic expression where $a = 1$, such as $x^2 - 5x + 6$?
2. What method would you use if the quadratic expression was not factorable by simple grouping?
3. Can you apply the same process if $c$ is positive instead of negative?
4. How do you recognize when a quadratic expression is prime (i.e., not factorable)?
5. What are some alternative methods for factoring quadratics, like completing the square?

Tip: When factoring quadratics, always look for two numbers that multiply to $a \times c$ and add to $b$ to make factoring easier.