Solving the Equation: (2x+3)(2x+5)(x-1)(x-2) = 30
This equation involves a product of four factors set equal to a constant. To solve it, we'll follow these steps:
1. Expanding the Equation
First, expand the product of the factors:
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Step 1: Expand the first two factors: (2x+3)(2x+5) = 4x² + 16x + 15
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Step 2: Expand the last two factors: (x-1)(x-2) = x² - 3x + 2
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Step 3: Combine the expansions: (4x² + 16x + 15)(x² - 3x + 2) = 30
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Step 4: Expand the entire product: 4x⁴ - 12x³ + 8x² + 16x³ - 48x² + 32x + 15x² - 45x + 30 = 30
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Step 5: Simplify by combining like terms: 4x⁴ + 4x³ - 25x² - 13x = 0
2. Finding the Solutions
Now we have a polynomial equation. To find its solutions, we can use factoring or the quadratic formula:
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Factoring: The equation can be factored as: x(4x³ + 4x² - 25x - 13) = 0
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Solution 1: One solution is immediately apparent: x = 0
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Finding other solutions: The remaining solutions must satisfy the cubic equation: 4x³ + 4x² - 25x - 13 = 0
This cubic equation doesn't easily factor. We can use numerical methods like the Rational Root Theorem or graphing to find approximate solutions.
Alternatively: We can use the Rational Root Theorem to test potential rational roots. This theorem states that any rational root of a polynomial with integer coefficients must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In this case, the constant term is -13 and the leading coefficient is 4. The factors of -13 are ±1 and ±13, and the factors of 4 are ±1, ±2, and ±4. Therefore, the possible rational roots are ±1, ±13, ±1/2, ±13/2, ±1/4, and ±13/4.
By testing these values, we can find the approximate solutions to the cubic equation.
3. Conclusion
The equation (2x+3)(2x+5)(x-1)(x-2) = 30 has four solutions:
- x = 0
- The other three solutions are found by solving the cubic equation 4x³ + 4x² - 25x - 13 = 0 using numerical methods or the Rational Root Theorem.
Note: The exact solutions to the cubic equation may be irrational numbers.