Given $$ f(x)=4 x^{3}+13 x^{2}-11 x-42 $$, answer the following. Part: $$ \mathbf{0} / \mathbf{2} $$ Part 1 of 2 Factor $$ f(x) $$, given that -2 is a zero. $$ f(x)= $$

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1. Since $$-2$$ is a zero of $$ f(x)=4x^3+13x^2-11x-42 $$, we know that $$(x+2)$$ is a factor.

2. We divide $$ f(x) $$ by $$(x+2)$$ using synthetic division. The coefficients of $$ f(x) $$ are: $$ 4 $$, $$ 13 $$, $$ -11 $$, and $$ -42 $$.

- Write $$-2$$ to the left and the coefficients in a row:
     
     $$
     \begin{array}{r|cccc}
     -2 & 4 & 13 & -11 & -42 \
       &   & -8 & -10 & 42 \
     \hline
       & 4 & 5 & -21 & 0 \
     \end{array}
     $$

- Steps:
     
     - Bring down $$4$$.
     - Multiply: $$-2 \times 4 = -8$$; add to $$13$$ to get $$5$$.
     - Multiply: $$-2 \times 5 = -10$$; add to $$-11$$ to get $$-21$$.
     - Multiply: $$-2 \times (-21) = 42$$; add to $$-42$$ to get $$0$$.

The remainder is $$0$$ and the quotient is $$ 4x^2+5x-21 $$.

3. Next, factor the quadratic $$ 4x^2+5x-21 $$.

- Look for two numbers whose product is $$ 4 \times (-21) = -84 $$ and whose sum is $$ 5 $$. These numbers are $$12$$ and $$-7$$.

- Rewrite the middle term:
     
     $$
     4x^2 + 12x - 7x - 21
     $$

- Group the terms:
     
     $$
     (4x^2 + 12x) + (-7x - 21)
     $$

- Factor each group:
     
     $$
     4x(x+3) - 7(x+3)
     $$

- Factor out the common binomial:
     
     $$
     (x+3)(4x-7)
     $$

4. Combine the factors from both steps:

$$
   f(x) = (x+2)(x+3)(4x-7)
   $$
$$ f(x) = (x + 2)(x + 3)(4x - 7) $$

Given \( f(x)=4 x^{3}+13 x^{2}-11 x-42 \), answer the following. Part: \( \mathbf{0} / \mathbf{2} \) Part 1 of 2 Factor \( f(x) \), given that -2 is a zero. \( f(x)= \)

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