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 for $$c$$ and $$d$$.
1. **State the problem:**
We have a polynomial $$Q(x) = x^3 + cx^2 + dx + 12$$.
We know:
- When divided by $$(x - 4)$$, remainder is 0.
- When divided by $$(x + 3)$$, remainder is 6.
We want to find $$c$$ and $$d$$ using the Remainder Theorem.
2. **Recall the Remainder Theorem:**
If a polynomial $$f(x)$$ is divided by $$(x - a)$$, the remainder is $$f(a)$$.
3. **Apply the Remainder Theorem:**
- For $$(x - 4)$$, remainder is 0, so:
$$Q(4) = 0$$
- For $$(x + 3)$$, remainder is 6, so:
$$Q(-3) = 6$$
4. **Write the equations:**
Calculate $$Q(4)$$:
$$Q(4) = 4^3 + c(4^2) + d(4) + 12 = 64 + 16c + 4d + 12 = 76 + 16c + 4d$$
Set equal to 0:
$$76 + 16c + 4d = 0$$
Calculate $$Q(-3)$$:
$$Q(-3) = (-3)^3 + c(-3)^2 + d(-3) + 12 = -27 + 9c - 3d + 12 = -15 + 9c - 3d$$
Set equal to 6:
$$-15 + 9c - 3d = 6$$
5. **Simplify the system:**
Equation 1:
$$16c + 4d = -76$$
Equation 2:
$$9c - 3d = 21$$
6. **Solve the system:**
Multiply Equation 2 by 4 to align $$d$$ terms:
$$36c - 12d = 84$$
Multiply Equation 1 by 3:
$$48c + 12d = -228$$
Add the two equations:
$$36c - 12d + 48c + 12d = 84 - 228$$
$$84c = -144$$
Solve for $$c$$:
$$c = \frac{-144}{84} = \frac{-12}{7}$$
7. **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 to clear denominator:
$$-192 + 28d = -532$$
Add 192 to both sides:
$$28d = -532 + 192 = -340$$
Divide both sides by 28:
$$d = \frac{-340}{28} = \frac{-85}{7}$$
**Final answer:**
$$c = -\frac{12}{7}, \quad d = -\frac{85}{7}$$