Correct Answer: (C) 1. Simplify the Expression: The given expression is $P = \frac{x^4+16x^2+64}{x^2}$. We can divide each term in the numerator by $x^2$ to rewrite the function: $P = \frac{x^4}{x^2} + \frac{16x^2}{x^2} + \frac{64}{x^2}$ $P = x^2 + 16 + \frac{64}{x^2}$ 2. Introduce a Substitution: Let $t = x^2$. Since $x$ is a non-zero real number, $x^2$ must be positive. Therefore, $t > 0$. Substituting $t$ into the simplified expression, we get: $P = t + 16 + \frac{64}{t}$ 3. Apply AM-GM Inequality: The Arithmetic Mean - Geometric Mean (AM-GM) inequality states that for any positive real numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$, which implies $a+b \geq 2\sqrt{ab}$. We apply this specifically to the variable terms $t$ and $\frac{64}{t}$: $t + \frac{64}{t} \geq 2\sqrt{t \cdot \frac{64}{t}}$ $t + \frac{64}{t} \geq 2\sqrt{64}$ $t + \frac{64}{t} \geq 2 \cdot 8 = 16$ 4. Determine the Minimum Value of P: From the AM-GM inequality, the minimum value of the variable component $t + \frac{64}{t}$ is 16. Substituting this back into the full expression for $P$: $P \geq 16 + 16$ $P \geq 32$ Thus, the minimum value of the expression is 32. 5. Condition for Equality: The equality in AM-GM holds when the terms are equal, i.e., $t = \frac{64}{t} \Rightarrow t^2 = 64$. Since $t > 0$, we have $t = 8$. As $t = x^2$, this occurs when $x = \pm\sqrt{8}$, confirming that a valid real $x$ exists for this minimum. Test Prep Tip: In CAT Algebra, whenever you encounter a rational expression where the numerator is a polynomial and the denominator is a single term (monomial), always perform polynomial division first. This often reveals a structure that allows for the immediate application of the AM-GM inequality for optimization.