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To solve this problem, we need to divide the polynomial \( 4x^3 + 2x + 1 \) by \( x + 1 \) using polynomial division.

### Steps for Polynomial Division
1. **Divide the first term of the dividend by the first term of the divisor**:
   - \( \frac{4x^3}{x} = 4x^2 \).

2. **Multiply the divisor by \( 4x^2 \)** and subtract:
   - \( (4x^2)(x + 1) = 4x^3 + 4x^2 \).
   - Subtract this result from the original polynomial:
     \[
     (4x^3 + 2x + 1) - (4x^3 + 4x^2) = -4x^2 + 2x + 1.
     \]

3. **Repeat the process with the new polynomial**:
   - Divide the first term, \( -4x^2 \), by \( x \):
     \[
     \frac{-4x^2}{x} = -4x.
     \]
   - Multiply \( x + 1 \) by \( -4x \):
     \[
     (-4x)(x + 1) = -4x^2 - 4x.
     \]
   - Subtract:
     \[
     (-4x^2 + 2x + 1) - (-4x^2 - 4x) = 6x + 1.
     \]

4. **Continue with the next term**:
   - Divide \( 6x \) by \( x \):
     \[
     \frac{6x}{x} = 6.
     \]
   - Multiply \( x + 1 \) by \( 6 \):
     \[
     6(x + 1) = 6x + 6.
     \]
   - Subtract:
     \[
     (6x + 1) - (6x + 6) = -5.
     \]

The final result of the division is:
\[
4x^2 - 4x + 6 - \frac{5}{x + 1}.
\]

### Answer
The correct answer is:
\[
\boxed{4x^2 - 4x + 6 - \frac{5}{x + 1}}
\]
which matches the fourth option in the list.

Would you like a breakdown of any specific step, or do you have any further questions?

Here are some related questions:
1. How do you use synthetic division for similar problems?
2. What happens if the divisor is not a binomial but a higher degree polynomial?
3. How can we verify the solution obtained from polynomial division?
4. How does the remainder theorem relate to polynomial division?
5. Can polynomial division be applied to non-polynomial expressions?

**Tip**: Polynomial division is very similar to long division with numbers; the main difference is that each term is handled separately based on its degree.

Learn how to divide the polynomial (4x^3 + 2x + 1) by (x + 1) using polynomial long division. This detailed step-by-step solution covers key algebraic concepts and is suitable for students in Grades 9-12.

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