Find $$ a $$ if $$ (x+2) $$ is a factor of $$ a x^{3}+4 x^{2}-4 $$

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To determine the value of $$ a $$ such that $$ (x + 2) $$ is a factor of the polynomial $$ P(x) = a x^{3} + 4 x^{2} - 4 $$, we can use the Factor Theorem. The Factor Theorem states that if $$ (x - c) $$ is a factor of $$ P(x) $$, then $$ P(c) = 0 $$.

Here, the factor is $$ (x + 2) $$, which can be rewritten as $$ (x - (-2)) $$. Therefore, we set $$ P(-2) = 0 $$:

$$
P(-2) = a (-2)^{3} + 4 (-2)^{2} - 4 = 0
$$

Calculating each term:

$$
(-2)^{3} = -8 \
4 (-2)^{2} = 4 \times 4 = 16 \
$$

Substitute these values into the equation:

$$
-8a + 16 - 4 = 0 \
-8a + 12 = 0 \
-8a = -12 \
a = \frac{12}{8} = \frac{3}{2}
$$

**Answer:** $$ \dfrac{3}{2} $$
$$ a = \frac{3}{2} $$

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