Write f(x) = x^3 - 4x^2 + 4x - 16 as a product of linear factors.

x = (x - 4)(x + 2i)(x - 2i) ?

looks good to me.

To factorize the given polynomial, we need to find its roots or zeros. The roots of a polynomial equation correspond to the values of x for which the polynomial is equal to zero. Once we have the roots, we can express the polynomial as a product of linear factors.

In this case, we are given the polynomial f(x) = x^3 - 4x^2 + 4x - 16. To find the roots, we set the polynomial equal to zero:

x^3 - 4x^2 + 4x - 16 = 0

Now, there are various methods to find the roots of a polynomial, such as factoring, synthetic division, or using numerical methods like the Newton-Raphson method. Since the polynomial is degree 3, it is manageable to use factoring or synthetic division.

However, upon examining the polynomial, it does not appear to have any obvious rational roots that can be found through factoring or synthetic division. In such cases, we can use the Rational Root Theorem to help us find a potential rational root before applying numerical methods.

The Rational Root Theorem states that if a polynomial has a rational root, it must be of the form ±(p/q), where p is a factor of the constant term, and q is a factor of the leading coefficient. In this case, the constant term is -16, and the leading coefficient is 1.

Factors of -16: ±1, ±2, ±4, ±8, ±16
Factors of 1: ±1

Therefore, the potential rational roots of the given polynomial are: ±1, ±2, ±4, ±8, ±16

To determine if any of these potential roots are actual roots of the polynomial, we substitute each value into the polynomial and see if it equals zero.

Substituting x = 1, we have:
f(1) = 1^3 - 4(1)^2 + 4(1) - 16 = 1 - 4 + 4 - 16 = -15 (not zero)

Substituting x = -1, we have:
f(-1) = (-1)^3 - 4(-1)^2 + 4(-1) - 16 = -1 - 4 - 4 - 16 = -25 (not zero)

Substituting x = 2, we have:
f(2) = 2^3 - 4(2)^2 + 4(2) - 16 = 8 - 16 + 8 - 16 = -16 (not zero)

Substituting x = -2, we have:
f(-2) = (-2)^3 - 4(-2)^2 + 4(-2) - 16 = -8 - 16 - 8 - 16 = -48 (not zero)

Substituting x = 4, we have:
f(4) = 4^3 - 4(4)^2 + 4(4) - 16 = 64 - 64 + 16 - 16 = 0

Therefore, the polynomial f(x) = x^3 - 4x^2 + 4x - 16 has a root at x = 4.

With this information, we can divide the polynomial by (x - 4) using either synthetic division or long division to obtain a quadratic quotient. The quotient will be a quadratic polynomial that can be further factored using methods like factoring by grouping or the quadratic formula.

Performing long division or synthetic division, we get:

(x^3 - 4x^2 + 4x - 16) / (x - 4) = x^2 + 4

Now, we have a quadratic polynomial, x^2 + 4, which does not have any real roots, but its roots are complex numbers. Specifically, the roots are imaginary numbers ±2i, where i is the imaginary unit, which is defined as the square root of -1.

Therefore, the factored form of f(x) = x^3 - 4x^2 + 4x - 16 is:

f(x) = (x - 4)(x + 2i)(x - 2i),

where (x + 2i) and (x - 2i) are the linear factors representing the complex roots.