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To solve the problem of factoring quadratic expressions, we will go through each section step by step. Factoring a quadratic expression \( ax^2 + bx + c \) involves finding two binomials whose product equals the given quadratic expression. Here's how we approach it:

Section A: Factoring Quadratic Expressions of the Form \( x^2 + bx + c \)

#### 1. \( x^2 + 7x - 30 \)
We need two numbers that multiply to \(-30\) and add to \(7\). These numbers are \(10\) and \(-3\).
\[
x^2 + 7x - 30 = (x + 10)(x - 3)
\]

#### 2. \( x^2 + 9x + 20 \)
We need two numbers that multiply to \(20\) and add to \(9\). These numbers are \(4\) and \(5\).
\[
x^2 + 9x + 20 = (x + 4)(x + 5)
\]

#### 3. \( x^2 + 8x - 9 \)
We need two numbers that multiply to \(-9\) and add to \(8\). These numbers are \(9\) and \(-1\).
\[
x^2 + 8x - 9 = (x + 9)(x - 1)
\]

#### 4. \( x^2 - 18x + 80 \)
We need two numbers that multiply to \(80\) and add to \(-18\). These numbers are \(-10\) and \(-8\).
\[
x^2 - 18x + 80 = (x - 10)(x - 8)
\]

#### 5. \( x^2 - 11x + 28 \)
We need two numbers that multiply to \(28\) and add to \(-11\). These numbers are \(-7\) and \(-4\).
\[
x^2 - 11x + 28 = (x - 7)(x - 4)
\]

#### 6. \( x^2 + 6x - 72 \)
We need two numbers that multiply to \(-72\) and add to \(6\). These numbers are \(12\) and \(-6\).
\[
x^2 + 6x - 72 = (x + 12)(x - 6)
\]

#### 7. \( x^2 - 9x - 22 \)
We need two numbers that multiply to \(-22\) and add to \(-9\). These numbers are \(-11\) and \(2\).
\[
x^2 - 9x - 22 = (x - 11)(x + 2)
\]

#### 8. \( x^2 - x - 12 \)
We need two numbers that multiply to \(-12\) and add to \(-1\). These numbers are \(-4\) and \(3\).
\[
x^2 - x - 12 = (x - 4)(x + 3)
\]

#### 9. \( x^2 + 3x - 108 \)
We need two numbers that multiply to \(-108\) and add to \(3\). These numbers are \(12\) and \(-9\).
\[
x^2 + 3x - 108 = (x + 12)(x - 9)
\]

#### 10. \( x^2 - 17x + 72 \)
We need two numbers that multiply to \(72\) and add to \(-17\). These numbers are \(-8\) and \(-9\).
\[
x^2 - 17x + 72 = (x - 8)(x - 9)
\]

#### 11. \( x^2 - x - 42 \)
We need two numbers that multiply to \(-42\) and add to \(-1\). These numbers are \(-7\) and \(6\).
\[
x^2 - x - 42 = (x - 7)(x + 6)
\]

#### 12. \( x^2 - 15x + 56 \)
We need two numbers that multiply to \(56\) and add to \(-15\). These numbers are \(-7\) and \(-8\).
\[
x^2 - 15x + 56 = (x - 7)(x - 8)
\]

Section B: Factoring Quadratic Expressions of the Form \( ax^2 + bx + c \)

#### 1. \( 2x^2 + 3x + 1 \)
We use the "ac method":
- \( ac = 2 \cdot 1 = 2 \)
- We need two numbers that multiply to \(2\) and add to \(3\). These numbers are \(2\) and \(1\).
- Rewrite the middle term: \( 2x^2 + 2x + x + 1 \)
- Factor by grouping: \( 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1) \)
\[
2x^2 + 3x + 1 = (2x + 1)(x + 1)
\]

#### 2. \( 2x^2 + 5x + 2 \)
We use the "ac method":
- \( ac = 2 \cdot 2 = 4 \)
- We need two numbers that multiply to \(4\) and add to \(5\). These numbers are \(4\) and \(1\).
- Rewrite the middle term: \( 2x^2 + 4x + x + 2 \)
- Factor by grouping: \( 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2) \)
\[
2x^2 + 5x + 2 = (2x + 1)(x + 2)
\]

#### 3. \( 2x^2 + 7x + 3 \)
We use the "ac method":
- \( ac = 2 \cdot 3 = 6 \)
- We need two numbers that multiply to \(6\) and add to \(7\). These numbers are \(6\) and \(1\).
- Rewrite the middle term: \( 2x^2 + 6x + x + 3 \)
- Factor by grouping: \( 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) \)
\[
2x^2 + 7x + 3 = (2x + 1)(x + 3)
\]

#### 4. \( 2x^2 + 7x + 5 \)
We use the "ac method":
- \( ac = 2 \cdot 5 = 10 \)
- We need two numbers that multiply to \(10\) and add to \(7\). These numbers are \(5\) and \(2\).
- Rewrite the middle term: \( 2x^2 + 5x + 2x + 5 \)
- Factor by grouping: \( x(2x + 5) + 1(2x + 5) = (x + 1)(2x + 5) \)
\[
2x^2 + 7x + 5 = (x + 1)(2x + 5)
\]

#### 5. \( 2x^2 + 9x + 7 \)
We use the "ac method":
- \( ac = 2 \cdot 7 = 14 \)
- We need two numbers that multiply to \(14\) and add to \(9\). These numbers are \(7\) and \(2\).
- Rewrite the middle term: \( 2x^2 + 7x + 2x + 7 \)
- Factor by grouping: \( x(2x + 7) + 1(2x + 7) = (x + 1)(2x + 7) \)
\[
2x^2 + 9x + 7 = (x + 1)(2x + 7)
\]

#### 6. \( 2x^2 + 5x + 3 \)
We use the "ac method":
- \( ac = 2 \cdot 3 = 6 \)
- We need two numbers that multiply to \(6\) and add to \(5\). These numbers are \(3\) and \(2\).
- Rewrite the middle term: \( 2x^2 + 3x + 2x + 3 \)
- Factor by grouping: \( x(2x + 3) + 1(2x + 3) = (x + 1)(2x + 3) \)
\[
2x^2 + 5x + 3 = (x + 1)(2x + 3)
\]

#### 7. \( 2x^2 + 8x + 6 \)
Factor out the GCF first:
\[
2(x^2 + 4x + 3)
\]
Now factor \( x^2 + 4x + 3 \):
- We need two numbers that multiply to \(3\) and add to \(4\). These numbers are \(3\) and \(1\).
\[
x^2 + 4x + 3 = (x + 3)(x + 1)
\]
So,
\[
2x^2 + 8x + 6 = 2(x + 3)(x + 1)
\]

#### 8. \( 2x^2 + 9x + 10 \)
We use the "ac method":
- \( ac = 2 \cdot 10 = 20 \)
- We need two numbers that multiply to \(20\) and add to \(9\). These numbers are \(5\) and \(4\).
- Rewrite the middle term: \( 2x^2 + 5x + 4x + 10 \)
- Factor by grouping: \( x(2x + 5) + 2(2x + 5) = (x + 2)(2x + 5) \)
\[
2x^2 + 9x + 10 = (x + 2)(2x + 5)
\]

#### 9. \( 2x^2 + 16x + 14 \)
Factor out the GCF first:
\[
2(x^2 + 8x + 7)
\]
Now factor \( x^2 + 8x + 7 \):
- We need two numbers that multiply to \(7\) and add to \(8\). These numbers are \(7\) and \(1\).
\[
x^2 + 8x + 7 = (x + 7)(x + 1)
\]
So,
\[
2x^2 + 16x + 14 = 2(x + 7)(x + 1)
\]

#### 10. \( 2x^2 + 16x + 24 \)
Factor out the GCF first:
\[
2(x^2 + 8x + 12)
\]
Now factor \( x^2 + 8x + 12 \):
- We need two numbers that multiply to \(12\) and add to \(8\). These numbers are \(6\) and \(2\).
\[
x^2 + 8x + 12 = (x + 6)(x + 2)
\]
So,
\[
2x^2 + 16x + 24 = 2(x + 6)(x + 2)
\]

#### 11. \( 2x^2 + 12x + 18 \)
Factor out the GCF first:
\[
2(x^2 + 6x + 9)
\]
Now factor \( x^2 + 6x + 9 \):
- This is a perfect square trinomial: \( (x + 3)^2 \)
\[
x^2 + 6x + 9 = (x + 3)^2
\]
So,
\[
2x^2 + 12x + 18 = 2(x + 3)^2
\]

#### 12. \( 2x^2 + 14x + 20 \)
Factor out the GCF first:
\[
2(x^2 + 7x + 10)
\]
Now factor \( x^2 + 7x + 10 \):
- We need two numbers that multiply to \(10\) and add to \(7\). These numbers are \(5\) and \(2\).
\[
x^2 + 7x + 10 = (x + 5)(x + 2)
\]
So,
\[
2x^2 + 14x + 20 = 2(x + 5)(x + 2)
\]

#### 13. \( 2x^2 + 22x + 36 \)
Factor out the GCF first:
\[
2(x^2 + 11x + 18)
\]
Now factor \( x^2 + 11x + 18 \):
- We need two numbers that multiply to \(18\) and add to \(11\). These numbers are \(9\) and \(2\).
\[
x^2 + 11x + 18 = (x + 9)(x + 2)
\]
So,
\[
2x^2 + 22x + 36 = 2(x + 9)(x + 2)
\]

#### 14. \( 2x^2 + 28x + 48 \)
Factor out the GCF first:
\[
2(x^2 + 14x + 24)
\]
Now factor \( x^2 + 14x + 24 \):
- We need two numbers that multiply to \(24\) and add to \(14\). These numbers are \(12\) and \(2\).
\[
x^2 + 14x + 24 = (x + 12)(x + 2)
\]
So,
\[
2x^2 + 28x + 48 = 2(x + 12)(x + 2)
\]

#### 15. \( 2x^2 + 26x + 72 \)
Factor out the GCF first:
\[
2(x^2 + 13x + 36)
\]
Now factor \( x^2 + 13x + 36 \):
- We need two numbers that multiply to \(36\) and add to \(13\). These numbers are \(9\) and \(4\).
\[
x^2 + 13x + 36 = (x + 9)(x + 4)
\]
So,
\[
2x^2 + 26x + 72 = 2(x + 9)(x + 4)
\]

Final Answer

\[
\boxed{
\begin{array}{lll}
\text{Section A:} & \begin{aligned}
1. & (x + 10)(x - 3) \\
2. & (x + 4)(x + 5) \\
3. & (x + 9)(x - 1) \\
4. & (x - 10)(x - 8) \\
5. & (x - 7)(x - 4) \\
6. & (x + 12)(x - 6) \\
7. & (x - 11)(x + 2) \\
8. & (x - 4)(x + 3) \\
9. & (x + 12)(x - 9) \\
10. & (x - 8)(x - 9) \\
11. & (x - 7)(x + 6) \\
12. & (x - 7)(x - 8)
\end{aligned} \\
\text{Section B:} & \begin{aligned}
1. & (2x + 1)(x + 1) \\
2. & (2x + 1)(x + 2) \\
3. & (2x + 1)(x + 3) \\
4. & (x + 1)(2x + 5) \\
5. & (x + 1)(2x + 7) \\
6. & (x + 1)(2x + 3) \\
7. & 2(x + 3)(x + 1) \\
8. & (x + 2)(2x + 5) \\
9. & 2(x + 7)(x + 1) \\
10. & 2(x + 6)(x + 2) \\
11. & 2(x + 3)^2 \\
12. & 2(x + 5)(x + 2) \\
13. & 2(x + 9)(x + 2) \\
14. & 2(x + 12)(x + 2) \\
15. & 2(x + 9)(x + 4)
\end{aligned}
\end{array}
}
\]