12. Find the values of $$ p $$ and $$ q $$ if both $$ (x-1) $$ and $$ (x+2) $$ are factors of the cubic $$ x^{3}+p x^{3}+q x+4 $$.

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Let's consider the cubic polynomial:

$$
P(x) = x^3 + p x^2 + q x + 4
$$

Given that both $$ (x - 1) $$ and $$ (x + 2) $$ are factors of $$ P(x) $$, it means that $$ x = 1 $$ and $$ x = -2 $$ are roots of the polynomial. Therefore, we can set up the following equations by substituting these values into $$ P(x) $$:

1. **For $$ x = 1 $$:**
$$
1^3 + p(1)^2 + q(1) + 4 = 0 \
1 + p + q + 4 = 0 \
p + q + 5 = 0 \quad \Rightarrow \quad p + q = -5 \quad \text{(Equation 1)}
$$

2. **For $$ x = -2 $$:**
$$
(-2)^3 + p(-2)^2 + q(-2) + 4 = 0 \
-8 + 4p - 2q + 4 = 0 \
-4 + 4p - 2q = 0 \quad \Rightarrow \quad 4p - 2q = 4 \quad \text{(Equation 2)}
$$

Now, we have a system of two equations:

$$
\begin{cases}
p + q = -5 \quad \text{(Equation 1)} \
4p - 2q = 4 \quad \text{(Equation 2)}
\end{cases}
$$

To solve this system, let's simplify Equation 2 by dividing all terms by 2:

$$
2p - q = 2 \quad \text{(Equation 3)}
$$

Now, add Equation 1 and Equation 3:

$$
\begin{align*}
(p + q) + (2p - q) &= -5 + 2 \
3p &= -3 \
p &= -1
\end{align*}
$$

Substitute $$ p = -1 $$ into Equation 1:

$$
-1 + q = -5 \
q = -4
$$

**Final Answer:**  
$$ p = -1 $$ and $$ q = -4 $$.
$$ p = -1 $$ and $$ q = -4 $$.

12. Find the values of \( p \) and \( q \) if both \( (x-1) \) and \( (x+2) \) are factors of the cubic \( x^{3}+p x^{3}+q x+4 \).

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