Key Concepts and Formulas

• Relationship Between Polynomial Roots and Values: For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ with roots $x_1, x_2, \dots, x_n$, we have $P(x) = a_n (x - x_1)(x - x_2)\dots(x - x_n)$. Thus, $P(k) = a_n \prod_{i=1}^n (k - x_i)$, which implies $\prod_{i=1}^n (k - x_i) = \frac{P(k)}{a_n}$.
• Sum of Squares Factorization: $A^2 + B^2 = (A - Bi)(A + Bi)$, where $i$ is the imaginary unit ($i^2 = -1$).
• Complex Conjugate Property: If a polynomial $P(x)$ has real coefficients, then for any complex number $z$, $P(\bar{z}) = \overline{P(z)}$.

Step-by-Step Solution

Step 1: Transform the Product Term $(x_i^2+4)$

We are given the expression $\prod_{i=1}^4 (4 + x_i^2)$. We want to factorize each term $4 + x_i^2$ using complex numbers so that we can relate it to the given polynomial. Using the sum of squares factorization, we have $4 + x_i^2 = 2^2 + x_i^2 = (2i - x_i)(-2i - x_i) = (x_i - 2i)(x_i + 2i)$. Therefore, $\prod_{i=1}^4 (4 + x_i^2) = \prod_{i=1}^4 (x_i - 2i)(x_i + 2i) = \left(\prod_{i=1}^4 (x_i - 2i)\right) \left(\prod_{i=1}^4 (x_i + 2i)\right)$.

Step 2: Connect the Product to Polynomial Evaluation

We want to rewrite the product in terms of the polynomial $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9$. We will use the relationship between polynomial roots and values. We can rewrite $\prod_{i=1}^4 (x_i - 2i)$ as $\prod_{i=1}^4 (-(2i - x_i)) = (-1)^4 \prod_{i=1}^4 (2i - x_i) = \prod_{i=1}^4 (2i - x_i)$. Using the formula $\prod_{i=1}^n (k - x_i) = \frac{P(k)}{a_n}$ with $k = 2i$, we have $\prod_{i=1}^4 (2i - x_i) = \frac{P(2i)}{a_4}$, where $a_4 = 4$ is the leading coefficient of the polynomial. Thus, $\prod_{i=1}^4 (x_i - 2i) = \frac{P(2i)}{4}$. Similarly, $\prod_{i=1}^4 (x_i + 2i) = \prod_{i=1}^4 (-(-2i) - x_i) = \prod_{i=1}^4 (-2i - x_i)$. Using the formula $\prod_{i=1}^n (k - x_i) = \frac{P(k)}{a_n}$ with $k = -2i$, we have $\prod_{i=1}^4 (-2i - x_i) = \frac{P(-2i)}{4}$. Thus, $\prod_{i=1}^4 (x_i + 2i) = \frac{P(-2i)}{4}$. Therefore, $\prod_{i=1}^4 (4 + x_i^2) = \frac{P(2i)}{4} \cdot \frac{P(-2i)}{4} = \frac{P(2i)P(-2i)}{16}$.

Step 3: Calculate $P(2i)$

We need to substitute $x = 2i$ into $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9$. $P(2i) = 4(2i)^4 + 8(2i)^3 - 17(2i)^2 - 12(2i) + 9 = 4(16) + 8(-8i) - 17(-4) - 12(2i) + 9 = 64 - 64i + 68 - 24i + 9 = 141 - 88i$.

Step 4: Calculate $P(-2i)$

Since $P(x)$ has real coefficients, we can use the property $P(\bar{z}) = \overline{P(z)}$. Here, $z = 2i$, so $\bar{z} = -2i$. Thus, $P(-2i) = \overline{P(2i)} = \overline{141 - 88i} = 141 + 88i$.

Step 5: Evaluate the Product $P(2i)P(-2i)$

We have $P(2i) = 141 - 88i$ and $P(-2i) = 141 + 88i$. So, $P(2i)P(-2i) = (141 - 88i)(141 + 88i) = 141^2 + 88^2 = 19881 + 7744 = 27625$. Therefore, $\prod_{i=1}^4 (4 + x_i^2) = \frac{P(2i)P(-2i)}{16} = \frac{27625}{16}$.

Step 6: Solving for $m$

We are given that $\prod_{i=1}^4 (4 + x_i^2) = \frac{125}{16} m$. We found that $\prod_{i=1}^4 (4 + x_i^2) = \frac{27625}{16}$. So, $\frac{27625}{16} = \frac{125}{16} m$. Multiplying both sides by $\frac{16}{125}$, we get $m = \frac{27625}{125} = 221$.

Common Mistakes & Tips

• Be careful with signs and powers of $i$ when evaluating the polynomial at complex numbers.
• Remember to use the complex conjugate property to simplify calculations when the polynomial has real coefficients.
• Do not forget to include the leading coefficient $a_n$ when using the relationship between polynomial roots and values.

Summary

We used the relationship between polynomial roots and values, along with complex number factorization and the complex conjugate property, to evaluate the product $\prod_{i=1}^4 (4 + x_i^2)$ as $\frac{27625}{16}$. Then, we used the given equation $\frac{27625}{16} = \frac{125}{16} m$ to solve for $m$, which resulted in $m = 221$.

Final Answer The final answer is \boxed{221}.