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### **Exercise 5**
We are given \( C = (5x - 3)^2 - (2x - 1)(5x - 3) \). Let's solve step-by-step:

#### 1. **Develop, reduce, and organize \( C \):**
- Expand each term:
  \[
  (5x - 3)^2 = 25x^2 - 30x + 9
  \]
  \[
  (2x - 1)(5x - 3) = 10x^2 - 11x + 3
  \]

- Subtract:
  \[
  C = (25x^2 - 30x + 9) - (10x^2 - 11x + 3)
  \]
  \[
  C = 25x^2 - 30x + 9 - 10x^2 + 11x - 3
  \]
  \[
  C = 15x^2 - 19x + 6
  \]

#### 2. **Calculate \( C \) for \( x = -2 \):**
Substitute \( x = -2 \) into \( C = 15x^2 - 19x + 6 \):
\[
C = 15(-2)^2 - 19(-2) + 6
\]
\[
C = 15(4) + 38 + 6 = 60 + 38 + 6 = 104
\]

#### 3. **Factorize \( C \):**
The quadratic \( C = 15x^2 - 19x + 6 \) factors as:
\[
C = (3x - 2)(5x - 3)
\]

#### 4. **Solve the equation \( (5x - 3)(2 - 3x) = 0 \):**
Set each factor to 0:
\[
5x - 3 = 0 \quad \text{or} \quad 2 - 3x = 0
\]
\[
x = \frac{3}{5} \quad \text{or} \quad x = \frac{2}{3}
\]

---

### **Exercise 6**
#### 1. **Simplify \( A, B, C \) without direct multiplication:**
- \( A = 109 \times 91 \):
  Use the difference of squares:
  \[
  109 \times 91 = (100 + 9)(100 - 9) = 100^2 - 9^2 = 10000 - 81 = 9919
  \]

- \( B = 125^2 - 75^2 \):
  Again, use the difference of squares:
  \[
  B = (125 - 75)(125 + 75) = 50 \times 200 = 10000
  \]

- \( C = 103^2 \):
  Use the formula \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( a = 100 \) and \( b = 3 \):
  \[
  C = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609
  \]

#### 2. **Calculate \( x(3x - 1) - 2(4x - 2) \) for \( x = -1 \):**
Substitute \( x = -1 \):
\[
x(3x - 1) - 2(4x - 2) = (-1)(3(-1) - 1) - 2(4(-1) - 2)
\]
\[
= (-1)(-3 - 1) - 2(-4 - 2) = (-1)(-4) - 2(-6)
\]
\[
= 4 + 12 = 16
\]

#### 3. **Calculate \( (3x - 4)^2 - (2x + 1)^2 \) for \( x = -2 \):**
Substitute \( x = -2 \):
\[
(3x - 4)^2 - (2x + 1)^2 = (3(-2) - 4)^2 - (2(-2) + 1)^2
\]
\[
= (-6 - 4)^2 - (-4 + 1)^2 = (-10)^2 - (-3)^2
\]
\[
= 100 - 9 = 91
\]

---

### **Exercise 7**
#### 1. **Express \( A_1 \) (area of rectangle \( ABCD \)) in terms of \( x \):**
The rectangle's dimensions are \( 2x + 1 \) (length) and \( 3x - 5 \) (width):
\[
A_1 = (2x + 1)(3x - 5)
\]

#### 2. **Express \( A_2 \) (area of polygon \( EFGHIJ \)) in terms of \( x \):**
The total area is made by adding the squares and subtracting the rectangle:
\[
A_2 = (3x - 7)^2 - z^2
\]
where \( z = (3x - 7) - (2x + 1) = x - 8 \), so \( z^2 = (x - 8)^2 \):
\[
A_2 = (3x - 7)^2 - (x - 8)^2
\]

Factorize using the difference of squares:
\[
A_2 = [(3x - 7) - (x - 8)][(3x - 7) + (x - 8)]
\]
\[
A_2 = (2x + 1)(4x - 15)
\]

#### 3. **Find \( x \) such that \( A_1 = A_2 \):**
Set \( (2x + 1)(3x - 5) = (2x + 1)(4x - 15) \). Divide by \( 2x + 1 \) (valid if \( x \neq -\frac{1}{2} \)):
\[
3x - 5 = 4x - 15
\]
\[
x = 10
\]

Verify that \( x = 10 \) satisfies the original expressions and is valid for the geometric conditions.

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### Let me know if you want further details or explanations! 

Here are 5 related questions for further practice:
1. How do you systematically expand and simplify algebraic expressions?
2. Can the area formulas in Exercise 7 be further simplified geometrically?
3. What is the significance of the difference of squares in simplifying calculations?
4. How do you check if solutions satisfy the conditions of a problem?
5. Why is factoring an essential skill in solving quadratic equations?

**Tip:** Always verify your final solutions by substituting back into the original problem to ensure consistency.

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