(x%5E3-1)(2x%2B3-x%5E2)-%3D0.jpeg "(x^3-27)(x^3-1)(2x+3-x^2) =0")
Factoring and Solving the Equation (x^3-27)(x^3-1)(2x+3-x^2) = 0
In this article, we will explore the process of factoring and solving the equation (x^3-27)(x^3-1)(2x+3-x^2) = 0.
Factoring the Equation
The given equation is a product of three binomials and a quadratic trinomial. To solve for x, we need to set each factor equal to zero and solve for x.
(x^3-27) = 0
This equation is a difference of cubes:
x^3 - 27 = x^3 - 3^3 = (x-3)(x^2+3x+3^2) = 0
This tells us that either (x-3) = 0 or (x^2+3x+3^2) = 0.
(x-3) = 0
Solving for x, we get:
x = 3
(x^2+3x+3^2) = 0
This quadratic equation cannot be factored further, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 3, and c = 3^2. Plugging these values into the formula, we get:
x = (-(3) ± √((3)^2 - 4(1)(3^2))) / 2(1)
Simplifying, we get two complex solutions:
x = (-3 ± √(-27)) / 2
(x^3-1) = 0
This equation is also a difference of cubes:
x^3 - 1 = x^3 - 1^3 = (x-1)(x^2+x+1) = 0
This tells us that either (x-1) = 0 or (x^2+x+1) = 0.
(x-1) = 0
Solving for x, we get:
x = 1
(x^2+x+1) = 0
This quadratic equation cannot be factored further, so we can use the quadratic formula again:
x = (-(1) ± √((1)^2 - 4(1)(1))) / 2(1)
Simplifying, we get two complex solutions:
x = (-1 ± √(-3)) / 2
(2x+3-x^2) = 0
This quadratic equation can be factored:
2x + 3 - x^2 = (x-1)(2+x) = 0
This tells us that either (x-1) = 0 or (2+x) = 0.
(x-1) = 0
We already know that x = 1 is a solution.
(2+x) = 0
Solving for x, we get:
x = -2
Solving the Equation
Combining all the solutions, we get:
x = 3 , x = 1 , x = -2 , x = (-3 ± √(-27)) / 2 , x = (-1 ± √(-3)) / 2
These are the roots of the equation (x^3-27)(x^3-1)(2x+3-x^2) = 0.
