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The problem requires dividing the polynomial \( x^3 + 3x^2 - 22x + 12 \) by \( x - 3 \) and, if there is a remainder, expressing it in the form \( \frac{r}{x - 3} \).

Let's perform polynomial division:

1. **Divide the leading term** of the dividend \( x^3 \) by the leading term of the divisor \( x \), which gives \( x^2 \).
   
2. **Multiply** \( x^2 \) by the entire divisor \( x - 3 \), resulting in \( x^3 - 3x^2 \).

3. **Subtract** this result from the original polynomial:
   \[
   (x^3 + 3x^2 - 22x + 12) - (x^3 - 3x^2) = 6x^2 - 22x + 12
   \]

4. **Repeat the process** with the new polynomial \( 6x^2 - 22x + 12 \):
   - Divide \( 6x^2 \) by \( x \) to get \( 6x \).
   - Multiply \( 6x \) by \( x - 3 \) to obtain \( 6x^2 - 18x \).
   - Subtract this from \( 6x^2 - 22x + 12 \):
     \[
     (6x^2 - 22x + 12) - (6x^2 - 18x) = -4x + 12
     \]

5. **Continue** with the remainder \( -4x + 12 \):
   - Divide \( -4x \) by \( x \) to get \( -4 \).
   - Multiply \( -4 \) by \( x - 3 \) to get \( -4x + 12 \).
   - Subtract this result:
     \[
     (-4x + 12) - (-4x + 12) = 0
     \]

Since there is no remainder, the division is exact, and the result is:
\[
x^2 + 6x - 4
\]

Thus, \( \frac{x^3 + 3x^2 - 22x + 12}{x - 3} = x^2 + 6x - 4 \).

Would you like additional details or have any questions about the process?

### Related Questions:
1. How do you handle remainders when dividing polynomials?
2. What are the steps to perform synthetic division?
3. How does polynomial division differ from long division with numbers?
4. Can all polynomials be divided without a remainder?
5. How can factoring help simplify polynomial division?

### Tip:
When dividing polynomials, align terms by degree to keep the process organized.

Divide. If there is a remainder, express it in the form \( \frac{r}{x - 3} \): \( \frac{x^3 + 3x^2 - 22x + 12}{x - 3} \)

Learn how to divide the polynomial \( x^3 + 3x^2 - 22x + 12 \) by \( x - 3 \) using polynomial division. Includes step-by-step calculations and handling of remainders. Suitable for students in Grades 10-12.

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