To solve this, let’s analyze each polynomial to see if it can be factored as a perfect square binomial.

1. Option (a) $x^2 + 4x + 64$:

   • To factor as a perfect square, the middle term should be $2 \times x \times \text{some number}$ such that we get 4 as the middle term.
   • However, this doesn’t satisfy the conditions of a perfect square because 4 is not twice the square root of 64.
   • Not a perfect square binomial.

2. Option (b) $x^2 - 64$:

   • This is a difference of squares, which factors as $(x - 8)(x + 8)$.
   • It is factorable but not as a perfect square binomial.
   • Not a perfect square binomial.

3. Option (c) $x^2 + 64$:

   • This expression is a sum of squares, which cannot be factored further over the real numbers.
   • Not a perfect square binomial.

4. Option (d) $x^2 + 16x + 64$:

   • This matches the form of a perfect square trinomial $(x + 8)^2$ because $16 = 2 \times x \times 8$ and $64 = 8^2$.
   • This is a perfect square binomial, as it factors to $(x + 8)^2$.

Answer: The correct option is (d) $x^2 + 16x + 64$.

Would you like a further breakdown of this process?

Here are some related questions:

1. What are the conditions for factoring a trinomial as a perfect square?
2. How do you identify a difference of squares?
3. Why can’t sums of squares be factored over real numbers?
4. Can a polynomial with an odd power be a perfect square?
5. What are some common factoring techniques for polynomials?

Tip: To determine if a trinomial is a perfect square, check if the first and last terms are squares and if the middle term is twice the product of their square roots.