Solving the Equation: (x^2+2x-8)(x^2-8x+15)=(x^2-x-20)(x^2+5x-14)
This equation involves a product of two quadratic expressions on each side. To solve it effectively, we can follow these steps:
1. Expand Both Sides of the Equation
First, expand the products on both sides of the equation using the distributive property or the FOIL method:
- Left-hand side:
- (x^2+2x-8)(x^2-8x+15) = x^4 - 6x^3 - 29x^2 + 126x - 120
- Right-hand side:
- (x^2-x-20)(x^2+5x-14) = x^4 + 4x^3 - 49x^2 - 60x + 280
2. Simplify the Equation
Now, we have a simplified equation:
x^4 - 6x^3 - 29x^2 + 126x - 120 = x^4 + 4x^3 - 49x^2 - 60x + 280
3. Combine Like Terms
Bring all the terms to one side of the equation:
-10x^3 + 20x^2 + 186x - 400 = 0
4. Factor the Equation
Now we need to factor the equation. This might seem tricky, but notice that all the coefficients are divisible by 2:
-5x^3 + 10x^2 + 93x - 200 = 0
We can factor this by grouping. The process involves finding a common factor among the first two terms and the last two terms.
-5x^3 + 10x^2 + 93x - 200 = 0 -5x^2(x - 2) + 93(x - 2) = 0 (-5x^2 + 93)(x - 2) = 0
Now, the equation is in factored form.
5. Solve for x
For the product of two factors to equal zero, at least one of them must be zero. This gives us two possible solutions:
- -5x^2 + 93 = 0
- 5x^2 = 93
- x^2 = 93/5
- x = ±√(93/5)
- x - 2 = 0
- x = 2
6. Solutions
Therefore, the solutions to the equation (x^2+2x-8)(x^2-8x+15)=(x^2-x-20)(x^2+5x-14) are:
- x = 2
- x = √(93/5)
- x = -√(93/5)
Note: It's always a good idea to check your solutions by substituting them back into the original equation to ensure they are valid.