1. Problem 1: Determine the other solution $w_2$ of $f(x)=x^2+2x+5$ given $w_1 = -1 + 2i$, and state the relationship between $w_1$ and $w_2$. 2. Formula and rule: For a polynomial with real coefficients nonreal complex roots occur in complex conjugate pairs. 3. For a quadratic $ax^2+bx+c$ the sum of the roots is $-b/a$ and the product of the roots is $c/a$. 4. Since $w_1 = -1 + 2i$ the complex conjugate is $\overline{w_1} = -1 - 2i$. 5. Therefore $w_2 = -1 - 2i$ and $w_2 = \overline{w_1}$. 6. Check: For $x^2+2x+5$ we have $a=1$, $b=2$, $c=5$. 7. Sum of roots is $-b/a = -2$ and product of roots is $c/a = 5$. 8. Compute $w_1 + w_2 = (-1 + 2i) + (-1 - 2i) = -2$ which matches the sum. 9. Compute $w_1 w_2 = (-1 + 2i)(-1 - 2i) = 5$ which matches the product. 10. Problem 2: Determine the modulus of $w_1 = -1 + 2i$. 11. Formula and rule: The modulus of a complex number $x + yi$ is $|x+yi| = \sqrt{x^2 + y^2}$. 12. Apply the formula: $|w_1| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$. 13. Problem 3: Determine all $x$ such that $|3x + 2| \le 5$. 14. Rule: For $|u| \le a$ with $a \ge 0$ we have $-a \le u \le a$. 15. Apply the rule: $-5 \le 3x + 2 \le 5$. 16. Subtract 2 from each part: $-7 \le 3x \le 3$. 17. Divide each part by 3: $-\frac{7}{3} \le x \le 1$. 18. Final answer for the inequality: $x \in [-\frac{7}{3},\,1]$. 19. Problem 4: Find the coefficient of $x$ in the expansion of $(5 - x/3)^3$. 20. Formula: The cubic binomial expansion is $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$. 21. Let $a=5$ and $b = -\frac{x}{3}$. 22. The linear term in $x$ comes from $3a^2 b$ which equals $3\cdot 5^2 \cdot (-\frac{x}{3})$. 23. Compute the linear term: $3\cdot 25 \cdot (-\frac{x}{3}) = 75 \cdot (-\frac{x}{3}) = -25x$. 24. Therefore the coefficient of $x$ is $-25$.