To find the zeros of the function $f(x) = x^2 - 8x + 16$, we set $f(x) = 0$ and solve for $x$:

$x^2 - 8x + 16 = 0$

This quadratic can be factored as follows:

$(x - 4)(x - 4) = 0$

or equivalently,

$(x - 4)^2 = 0$

To find the solutions, we set $x - 4 = 0$:

$x = 4$

Thus, the only zero of $f(x) = x^2 - 8x + 16$ is:

$x = 4$

This is a repeated root, meaning $x = 4$ is a solution with multiplicity 2.

Would you like more details or have any questions?

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Here are some related questions that expand on this topic:

1. What is the geometric interpretation of a repeated root for a quadratic function?
2. How would you find the vertex of the function $f(x) = x^2 - 8x + 16$?
3. How would the zeros change if the function were $f(x) = x^2 - 8x + 15$?
4. Can we use the quadratic formula to verify the result, and how?
5. How can we complete the square to confirm the form $(x - 4)^2$?

Tip: For any quadratic equation of the form $ax^2 + bx + c = 0$, if the discriminant $b^2 - 4ac = 0$, the quadratic has a repeated root.