This answer key provides the correct quotients and remainders for four polynomial long division practice problems.

The task involves performing polynomial long division for each given problem and determining the quotient and remainder. Below, I will explain the solution for each problem step by step.

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Problem 1:

Divide \( x^2 - 3x - 6 \) by \( x + 2 \).

#### Step-by-Step Solution:
1. Set up the division:
\[
\begin{array}{r|rr}
x + 2 & x^2 - 3x - 6 \\
\end{array}
\]

2. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{x^2}{x} = x
\]
Write \( x \) above the division bar.

3. Multiply \( x \) by the divisor \( x + 2 \):
\[
x \cdot (x + 2) = x^2 + 2x
\]
Write this below the dividend.

4. Subtract:
\[
(x^2 - 3x - 6) - (x^2 + 2x) = -5x - 6
\]

5. Repeat the process with the new polynomial \( -5x - 6 \):
\[
\frac{-5x}{x} = -5
\]
Write \( -5 \) above the division bar next to \( x \).

6. Multiply \( -5 \) by the divisor \( x + 2 \):
\[
-5 \cdot (x + 2) = -5x - 10
\]
Write this below \( -5x - 6 \).

7. Subtract:
\[
(-5x - 6) - (-5x - 10) = 4
\]

8. Final result:
The quotient is \( x - 5 \) and the remainder is \( 4 \).

\[
\boxed{x - 5 \text{ Remainder } 4}
\]

---

Problem 2:

Divide \( x^3 - 4x - 8 \) by \( x - 3 \).

#### Step-by-Step Solution:
1. Set up the division:
\[
\begin{array}{r|rrr}
x - 3 & x^3 & 0x^2 & -4x & -8 \\
\end{array}
\]
Note: We include a placeholder \( 0x^2 \) since there is no \( x^2 \)-term in the dividend.

2. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{x^3}{x} = x^2
\]
Write \( x^2 \) above the division bar.

3. Multiply \( x^2 \) by the divisor \( x - 3 \):
\[
x^2 \cdot (x - 3) = x^3 - 3x^2
\]
Write this below the dividend.

4. Subtract:
\[
(x^3 + 0x^2 - 4x - 8) - (x^3 - 3x^2) = 3x^2 - 4x - 8
\]

5. Repeat the process with the new polynomial \( 3x^2 - 4x - 8 \):
\[
\frac{3x^2}{x} = 3x
\]
Write \( 3x \) above the division bar next to \( x^2 \).

6. Multiply \( 3x \) by the divisor \( x - 3 \):
\[
3x \cdot (x - 3) = 3x^2 - 9x
\]
Write this below \( 3x^2 - 4x - 8 \).

7. Subtract:
\[
(3x^2 - 4x - 8) - (3x^2 - 9x) = 5x - 8
\]

8. Repeat the process with the new polynomial \( 5x - 8 \):
\[
\frac{5x}{x} = 5
\]
Write \( 5 \) above the division bar next to \( 3x \).

9. Multiply \( 5 \) by the divisor \( x - 3 \):
\[
5 \cdot (x - 3) = 5x - 15
\]
Write this below \( 5x - 8 \).

10. Subtract:
\[
(5x - 8) - (5x - 15) = 7
\]

11. Final result:
The quotient is \( x^2 + 3x + 5 \) and the remainder is \( 7 \).

\[
\boxed{x^2 + 3x + 5 \text{ Remainder } 7}
\]

---

Problem 3:

Divide \( 2x^2 - 6x - 12 \) by \( x - 3 \).

#### Step-by-Step Solution:
1. Set up the division:
\[
\begin{array}{r|rrr}
x - 3 & 2x^2 & -6x & -12 \\
\end{array}
\]

2. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{2x^2}{x} = 2x
\]
Write \( 2x \) above the division bar.

3. Multiply \( 2x \) by the divisor \( x - 3 \):
\[
2x \cdot (x - 3) = 2x^2 - 6x
\]
Write this below the dividend.

4. Subtract:
\[
(2x^2 - 6x - 12) - (2x^2 - 6x) = -12
\]

5. Final result:
The quotient is \( 2x \) and the remainder is \( -12 \).

\[
\boxed{2x \text{ Remainder } -12}
\]

---

Problem 4:

Divide \( 6x^2 + 5x - 3 \) by \( 2x - 1 \).

#### Step-by-Step Solution:
1. Set up the division:
\[
\begin{array}{r|rrr}
2x - 1 & 6x^2 & 5x & -3 \\
\end{array}
\]

2. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{6x^2}{2x} = 3x
\]
Write \( 3x \) above the division bar.

3. Multiply \( 3x \) by the divisor \( 2x - 1 \):
\[
3x \cdot (2x - 1) = 6x^2 - 3x
\]
Write this below the dividend.

4. Subtract:
\[
(6x^2 + 5x - 3) - (6x^2 - 3x) = 8x - 3
\]

5. Repeat the process with the new polynomial \( 8x - 3 \):
\[
\frac{8x}{2x} = 4
\]
Write \( 4 \) above the division bar next to \( 3x \).

6. Multiply \( 4 \) by the divisor \( 2x - 1 \):
\[
4 \cdot (2x - 1) = 8x - 4
\]
Write this below \( 8x - 3 \).

7. Subtract:
\[
(8x - 3) - (8x - 4) = 1
\]

8. Final result:
The quotient is \( 3x + 4 \) and the remainder is \( 1 \).

\[
\boxed{3x + 4 \text{ Remainder } 1}
\]

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Final Answers:

1. \( x - 5 \text{ Remainder } 4 \)
2. \( x^2 + 3x + 5 \text{ Remainder } 7 \)
3. \( 2x \text{ Remainder } -12 \)
4. \( 3x + 4 \text{ Remainder } 1 \)

\[
\boxed{x - 5 \text{ Remainder } 4, \quad x^2 + 3x + 5 \text{ Remainder } 7, \quad 2x \text{ Remainder } -12, \quad 3x + 4 \text{ Remainder } 1}
\]