1. **State the problem:** Simplify or analyze the quadratic expression $x^2 + 10x + 41$. 2. **Recall the quadratic form:** A quadratic expression is generally written as $ax^2 + bx + c$. 3. **Identify coefficients:** Here, $a=1$, $b=10$, and $c=41$. 4. **Check if it can be factored easily:** We look for two numbers that multiply to $41$ and add to $10$. Since $41$ is prime and no such integer pair exists, it cannot be factored nicely over the integers. 5. **Complete the square:** To rewrite in vertex form, use the formula: $$x^2 + bx + c = (x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2 + c$$ 6. **Apply the formula:** $$x^2 + 10x + 41 = \left(x + \frac{10}{2}\right)^2 - \left(\frac{10}{2}\right)^2 + 41 = (x + 5)^2 - 25 + 41$$ 7. **Simplify:** $$ (x + 5)^2 + 16 $$ 8. **Interpretation:** The expression is a perfect square plus 16, so its minimum value is 16 when $x = -5$. **Final answer:** $$x^2 + 10x + 41 = (x + 5)^2 + 16$$