Factor the following trinomial. $$ \begin{array}{l}9 x^{2}+48 x+64 \ ([?] x+\square)^{2}\end{array} $$

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To factor the trinomial $$ 9x^{2} + 48x + 64 $$ as a perfect square, we can follow these steps:

1. **Identify the Structure of a Perfect Square Trinomial:**

A perfect square trinomial has the form:
   $$
   (ax + b)^2 = a^2x^2 + 2abx + b^2
   $$

2. **Match the Coefficients:**
   
   Compare $$ 9x^{2} + 48x + 64 $$ with $$ a^2x^2 + 2abx + b^2 $$:
   
   - **For $$ a^2x^2 = 9x^2 $$:**
     $$
     a^2 = 9 \implies a = 3
     $$
     
   - **For $$ 2abx = 48x $$:**
     $$
     2 \times 3 \times b = 48 \implies 6b = 48 \implies b = 8
     $$
     
   - **Verify $$ b^2 = 64 $$:**
     $$
     8^2 = 64 \quad \text{(which matches)}
     $$
   
3. **Write the Factored Form:**

Substituting $$ a = 3 $$ and $$ b = 8 $$ into the perfect square form:
   $$
   (3x + 8)^2
   $$

**Therefore, the factored form of the trinomial is:**
$$
(3x + 8)^2
$$
The trinomial $$ 9x^{2} + 48x + 64 $$ factors to $$ (3x + 8)^2 $$.

Factor the following trinomial.

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