find all zeros of the following function:

f(x)=x^3-x^2+4x-4
I have (x^2+4)(x+1)

so one zero is 1. how do i find the two other zeros.

if +1 is a zero, (x-1) is a factor because (x-1) is zero when x = 1

I have

(x^2+4)(x-1)
then when is (x^2 + 0 x +4 )= 0??
x = [ 0 +/- sqrt (0 -16) ] / 2
x = +/- 2 i (imaginary roots)

To find the remaining zeros of the function f(x) = x^3 - x^2 + 4x - 4, we can use the factor theorem and synthetic division.

Step 1: Start with the polynomial expression (x^2 + 4)(x + 1). Since we already found one zero to be 1, we can continue with the factor (x - 1).

Step 2: Apply synthetic division to divide f(x) by (x - 1). Write down the coefficients of the terms in f(x) in order of descending powers of x. Then, bring down the coefficient of the highest power term (1 in this case) as shown below:

1 | 1 -1 4 -4
| 1 0 4
________________
1 0 4 0

The result of synthetic division gives us a quotient of x^2 + 4 and a remainder of 0. The values on the bottom row represent the coefficients of the reduced polynomial.

Step 3: Write the reduced polynomial. Since the remainder is 0, the quotient x^2 + 4 represents the remaining factors of f(x). Therefore, the fully factored form is:

f(x) = (x - 1)(x^2 + 4)

Now, to find the remaining zeros, we need to solve the equation x^2 + 4 = 0.

Step 4: Solve x^2 + 4 = 0. Subtracting 4 from both sides gives:

x^2 = -4

Since the equation has no real solutions (the square of a real number cannot be negative), the remaining zeros of the function do not exist in the real number system. However, if we consider complex numbers, the solutions are x = ± 2i, where i represents the imaginary unit (√-1).

Therefore, the zeros of f(x) = x^3 - x^2 + 4x - 4 are x = 1, x = 2i, and x = -2i.