## Solving the Equation (2x+3)(2x+5)(x-1)(x-2) = 30

This equation involves a product of four linear expressions, making it seem quite complex. However, we can solve it by employing some strategic manipulation and algebraic techniques.

### Step 1: Expand the Product

Firstly, we expand the product of the four expressions:

(2x + 3)(2x + 5)(x - 1)(x - 2) = 30

Expanding the first two terms and the last two terms separately, we get:

(4x² + 16x + 15)(x² - 3x + 2) = 30

### Step 2: Simplify and Rearrange

Now, we multiply the two quadratic expressions and rearrange the terms to get a polynomial equation:

4x⁴ - 4x³ - 29x² + 54x + 30 = 30

Subtracting 30 from both sides, we obtain:

4x⁴ - 4x³ - 29x² + 54x = 0

### Step 3: Factor out a Common Factor

We can factor out a common factor of 'x' from the left-hand side:

x(4x³ - 4x² - 29x + 54) = 0

### Step 4: Factor the Cubic Expression

The cubic expression inside the parentheses can be factored using various techniques, such as grouping or the Rational Root Theorem. For this example, we will use grouping:

x[(4x³ - 4x²) + (-29x + 54)] = 0

x[4x²(x - 1) - 29(x - 1)] = 0

x(x - 1)(4x² - 29) = 0

### Step 5: Find the Solutions

We have now expressed the equation as a product of three factors. For the product to be zero, at least one of the factors must be zero. This leads to three possible solutions:

**x = 0****x = 1****4x² - 29 = 0**

Solving the last equation for x:

4x² = 29

x² = 29/4

**x = ±√(29/4)**

Therefore, the solutions to the equation (2x+3)(2x+5)(x-1)(x-2) = 30 are:

**x = 0, x = 1, x = √(29/4), x = -√(29/4)**