Question: Consider a new polynomial function $Q(x) = x^3 + cx^2 + dx + 12$. When $Q(x)$ is divided by $(x - 4)$, it leaves a remainder of $0$. When $Q(x)$ is divided by $(x + 3)$, it leaves a remainder of $6$. Use the Remainder Theorem to form two equations in terms of $c$ and $d$. Then, use your scientific calculator to solve the system.
1. **State the problem:**
We have a polynomial $Q(x) = x^3 + cx^2 + dx + 12$.
2. **Use the Remainder Theorem:**
- When divided by $(x - 4)$, remainder is $0$, so $Q(4) = 0$.
- When divided by $(x + 3)$, remainder is $6$, so $Q(-3) = 6$.
3. **Form equations:**
Calculate $Q(4)$:
$$Q(4) = 4^3 + c(4^2) + d(4) + 12 = 64 + 16c + 4d + 12 = 76 + 16c + 4d$$
Set equal to zero:
$$76 + 16c + 4d = 0$$
Calculate $Q(-3)$:
$$Q(-3) = (-3)^3 + c(-3)^2 + d(-3) + 12 = -27 + 9c - 3d + 12 = -15 + 9c - 3d$$
Set equal to 6:
$$-15 + 9c - 3d = 6$$
4. **Simplify the system:**
Equation 1:
$$16c + 4d = -76$$
Equation 2:
$$9c - 3d = 21$$
5. **Solve the system:**
Multiply Equation 1 by 3:
$$48c + 12d = -228$$
Multiply Equation 2 by 4:
$$36c - 12d = 84$$
Add the two equations:
$$48c + 12d + 36c - 12d = -228 + 84$$
$$84c = -144$$
$$c = \frac{-144}{84} = \frac{-12}{7}$$
6. **Find $d$:**
Substitute $c = -\frac{12}{7}$ into Equation 1:
$$16\left(-\frac{12}{7}\right) + 4d = -76$$
$$-\frac{192}{7} + 4d = -76$$
Multiply both sides by 7:
$$-192 + 28d = -532$$
$$28d = -532 + 192 = -340$$
$$d = \frac{-340}{28} = \frac{-85}{7}$$
**Final answer:**
$$c = -\frac{12}{7}, \quad d = -\frac{85}{7}$$