Evaluate Fourth Degree Polynomial with Given Extrema and Limit

To find the value of $p(2)$, we begin by analyzing the given limit:
$\lim_{x \to 0} \left( 1 + \frac{p(x)}{x^2} \right) = 2$

Subtracting $1$ from both sides gives:
$\lim_{x \to 0} \frac{p(x)}{x^2} = 1$

Since $p(x)$ is a polynomial of degree 4, we can express it in standard form as:
$p(x) = ax^4 + bx^3 + cx^2 + dx + e$

For the limit $\lim_{x \to 0} \frac{p(x)}{x^2}$ to exist and be finite, the terms of degree less than 2 must be zero:
$e = 0 \quad \text{and} \quad d = 0$

Furthermore, the coefficient of $x^2$ is given by the value of the limit:
$c = 1$

Thus, the polynomial takes the form:
$p(x) = ax^4 + bx^3 + x^2$

Next, we calculate the first derivative of $p(x)$:
$p'(x) = 4ax^3 + 3bx^2 + 2x$

Since $p(x)$ has extrema at $x = 1$ and $x = 2$, we have $p'(1) = 0$ and $p'(2) = 0$:

1. For $x = 1$:
   $p'(1) = 4a(1)^3 + 3b(1)^2 + 2(1) = 0 \implies 4a + 3b + 2 = 0 \implies 4a + 3b = -2$

2. For $x = 2$:
   $p'(2) = 4a(2)^3 + 3b(2)^2 + 2(2) = 0 \implies 32a + 12b + 4 = 0$
   Dividing this equation by 4 gives:
   $8a + 3b = -1$

Now, subtracting the first equation from the second equation:
$(8a + 3b) - (4a + 3b) = -1 - (-2)$
$4a = 1 \implies a = \frac{1}{4}$

Substituting $a = \frac{1}{4}$ back into $4a + 3b = -2$:
$4\left(\frac{1}{4}\right) + 3b = -2 \implies 1 + 3b = -2 \implies 3b = -3 \implies b = -1$

Therefore, the polynomial is:
$p(x) = \frac{1}{4}x^4 - x^3 + x^2$

Finally, we evaluate $p(2)$:
$p(2) = \frac{1}{4}(2)^4 - (2)^3 + (2)^2 = \frac{1}{4}(16) - 8 + 4 = 4 - 8 + 4 = 0$