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To solve this problem, we need to divide the polynomial \(3x^3 + 2x^2 - 3x + 54\) by \(x + 3\) using polynomial long division.

### Steps for Polynomial Long Division:
1. **Divide the leading term of the dividend by the leading term of the divisor**:
   - \( \frac{3x^3}{x} = 3x^2 \)
   - This becomes the first term in our quotient.

2. **Multiply the entire divisor by this term**:
   - \(3x^2 \cdot (x + 3) = 3x^3 + 9x^2\)

3. **Subtract this result from the original polynomial**:
   - \((3x^3 + 2x^2 - 3x + 54) - (3x^3 + 9x^2) = -7x^2 - 3x + 54\)

4. **Repeat the process** with the new polynomial \(-7x^2 - 3x + 54\):
   - Divide \(-7x^2\) by \(x\): \(-7x\)
   - Multiply \(-7x\) by \(x + 3\): \(-7x^2 - 21x\)
   - Subtract: \((-7x^2 - 3x + 54) - (-7x^2 - 21x) = 18x + 54\)

5. **Continue the process** with the remaining polynomial \(18x + 54\):
   - Divide \(18x\) by \(x\): \(18\)
   - Multiply \(18\) by \(x + 3\): \(18x + 54\)
   - Subtract: \((18x + 54) - (18x + 54) = 0\)

Since there is no remainder, the division is complete.

### Quotient
The quotient is:
\[
3x^2 - 7x + 18
\]

So the completed answer is:
\[
3x^2 - 7x + 18
\]

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Would you like further details on any step or have any questions?

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Here are five related questions to deepen your understanding:

1. How would you perform polynomial division if there was a remainder?
2. What is synthetic division, and how does it differ from polynomial long division?
3. Can you divide polynomials when the divisor is of a higher degree than the dividend?
4. How can you check if your polynomial division answer is correct?
5. What are some practical applications of polynomial division?

**Tip:** When dividing polynomials, always align terms by their degree to avoid mistakes during subtraction.

Divide \( \frac{3x^3 + 2x^2 - 3x + 54}{x + 3} \) and find the quotient.

Learn how to divide the polynomial 3x^3 + 2x^2 - 3x + 54 by x + 3 using polynomial long division. This solution includes a step-by-step guide, covering key algebraic concepts in polynomial division, suitable for high school students.

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