The function f(x) has three zeros: -2, (1 + √(3)i) / 2, and (1 - √(3)i) / 2. These solutions include an imaginary component due to the square root of -3.
Let's find all the zeros of the function:
f(x) = 4x^3 + 4x^2 - 4x + 8
Step 1: Factoring the polynomial
We can see that 4 is a common factor for all terms. Factoring it out and rearranging, we get:
f(x) = 4(x^3 + x^2 - x + 2)
Now, we can factor the remaining expression within the parentheses using trial and error or other methods. We find that it factors into:
f(x) = 4(x + 2)(x^2 - x + 1)
Step 2: Solving the factored equation
The expression is equal to zero when either of the factored parts is equal to zero. So, we have two equations to solve:
x + 2 = 0 and x^2 - x + 1 = 0
For the first equation, we simply solve for x:
x = -2
For the second equation, this is a quadratic equation. We can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are the coefficients of the equation. In this case, a = 1, b = -1, and c = 1. Substituting these values, we get:
x = (1 ± √((-1)^2 - 4 * 1 * 1)) / (2 * 1)
x = (1 ± √(-3)) / 2
Unfortunately, the square root of -3 is an imaginary number, denoted by "i". So, the solutions will involve "i":
x = (1 ± √(3)i) / 2
Step 3: Answer
Therefore, the zeros of f(x) are:
x = -2
x = (1 + √(3)i) / 2
x = (1 - √(3)i) / 2
Complete question:
Find all zeros of f(x) = 4x^3 + 4x^2 - 4x + 8. Not decimal approximation.