Here are the step-by-step solutions for the problems. 1. Factorize $x^2 + 3x + 2$ Step 1: Find two numbers that multiply to the constant term (2) and add up to the coefficient of the middle term (3). The numbers are 1 and 2, because $1 \times 2 = 2$ and $1 + 2 = 3$. Step 2: Write the quadratic expression in factored form using these numbers. $$x^2 + 3x + 2 = (x + 1)(x + 2)$$ The factored form is $\boxed{(x + 1)(x + 2)}$. 2. Solve the equation $(x+1)(x+2)$ Assuming the equation is $(x+1)(x+2) = 0$, as is standard for "solve the equation" when given factors. Step 1: Set each factor equal to zero. For a product of factors to be zero, at least one of the factors must be zero. $$x + 1 = 0 \quad \text{or} \quad x + 2 = 0$$ Step 2: Solve each linear equation for x. $$x + 1 = 0 \implies x = -1$$ $$x + 2 = 0 \implies x = -2$$ The solutions are $\boxed{x = -1, x = -2}$. 3. Solve for x: $5x \ge 2y$ Where $x=6, y=2$ This question asks to "solve for x" but then provides a specific value for x. We will first solve the inequality for x in terms of y, and then verify if the given values satisfy the inequality. Step 1: Solve the inequality for x. $$5x \ge 2y$$ Divide both sides by 5: $$x \ge \frac{2y}{5}$$ Step 2: Substitute the given values $x=6$ and $y=2$ into the original inequality to verify. $$5(6) \ge 2(2)$$ $$30 \ge 4$$ This statement is true. The solution for x in terms of y is $\boxed{x \ge \frac{2y}{5}}$. The given values $x=6, y=2$ satisfy the inequality. 4. Solve for a by dividing the equation $5a + 25 \le 10$ by -5 Step 1: Write down the inequality. $$5a + 25 \le 10$$ Step 2: Divide every term in the inequality by -5. Remember to reverse the inequality sign when dividing by a negative number. $$\frac{5a}{-5} + \frac{25}{-5} \ge \frac{10}{-5}$$ $$-a - 5 \ge -2$$ Step 3: Isolate the term with 'a' by adding 5 to both sides. $$-a \ge -2 + 5$$ $$-a \ge 3$$ Step 4: Multiply both sides by -1 to solve for 'a'. Remember to reverse the inequality sign again. $$(-1)(-a) \le (-1)(3)$$ $$a \le -3$$ The solution is $\boxed{a \le -3}$.