(x%5E2-3x%2B9)-x(x-2)(x%2B2)%2B11%3D0.jpeg "(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0")
Solving the Equation: (x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0
This equation might look intimidating at first, but with a few algebraic manipulations, it can be solved. Let's break down the steps:
Step 1: Expand the Products
First, we need to expand the products in the equation using the distributive property and the difference of squares formula:
- (x+3)(x^2-3x+9) This is a special product known as the sum of cubes: (a+b)(a^2-ab+b^2) = a^3 + b^3. Applying this, we get x^3 + 27.
- x(x-2)(x+2) This is the product of a difference of squares: (a-b)(a+b) = a^2 - b^2. Expanding, we get x(x^2 - 4) = x^3 - 4x.
Step 2: Simplify the Equation
Now, substitute the expanded forms back into the original equation:
x^3 + 27 - (x^3 - 4x) + 11 = 0
Simplifying further:
x^3 + 27 - x^3 + 4x + 11 = 0
Combining like terms:
4x + 38 = 0
Step 3: Solve for x
Finally, we can solve for x by isolating it:
4x = -38
x = -38/4
x = -9.5
Conclusion
Therefore, the solution to the equation (x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0 is x = -9.5.