If (x-1) is a factor of the polynomial f(x)=4x³−4x²−x−k Factorize completely and state it's zeros. Do this without leaving out any details

If $(x-1)$ is a factor of $f(x)$, then $f(1) = 0$.
Substitute $x=1$ into the polynomial $f(x) = 4x^3 - 4x^2 - x - k$:

The polynomial becomes:

Since $(x-1)$ is a factor, we divide $f(x)$ by $(x-1)$.
Using synthetic division with the root $x=1$:

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

The quotient is $4x^2 + 0x - 1$, which simplifies to $4x^2 - 1$.
So, $f(x) = (x-1)(4x^2 - 1)$.

The quadratic expression $4x^2 - 1$ is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=2x$ and $b=1$.

Substitute the factored quadratic back into the expression for $f(x)$:

The completely factorized polynomial is $\boxed{f(x) = (x-1)(2x-1)(2x+1)}$.

To find the zeros, set each factor equal to zero and solve for $x$:
For the first factor:

For the second factor:

For the third factor:

The zeros of the polynomial are $\boxed{1, \frac{1}{2}, -\frac{1}{2}}$.