2. Solve each equation by rewriting it in factored form and using the zero prod property. Show your reasoning. a. \( x^{2}+8 x+15=0 \) b. \( x^{2}-8 x+12=5 \) c. \( x^{2}-10 x-11=0 \) d. \( 49-x^{2}=0 \) e. \( (x+4)(x+5)-30=0 \)

Sure, let's solve each equation by factoring and applying the **Zero Product Property**, which states that if the product of two factors is zero, then at least one of the factors must be zero. --- ### **Problem 2a.** \[ x^{2} + 8x + 15 = 0 \] **Step 1: Factor the quadratic equation.** We need to find two numbers that multiply to **15** (the constant term) and add up to **8** (the coefficient of \( x \)). - **3 and 5** satisfy this condition because: \[ 3 \times 5 = 15 \] \[ 3 + 5 = 8 \] **Step 2: Rewrite the equation in its factored form.** \[ (x + 3)(x + 5) = 0 \] **Step 3: Apply the Zero Product Property.** Set each factor equal to zero: 1. \( x + 3 = 0 \) \[ x = -3 \] 2. \( x + 5 = 0 \) \[ x = -5 \] **Solution:** \[ x = -3 \quad \text{and} \quad x = -5 \] --- ### **Problem 2b.** \[ x^{2} - 8x + 12 = 5 \] **Step 1: Move all terms to one side to set the equation to zero.** \[ x^{2} - 8x + 12 - 5 = 0 \] \[ x^{2} - 8x + 7 = 0 \] **Step 2: Factor the quadratic equation.** We need two numbers that multiply to **7** and add up to **-8**. - **-1 and -7** satisfy this condition because: \[ (-1) \times (-7) = 7 \] \[ (-1) + (-7) = -8 \] **Step 3: Rewrite the equation in its factored form.** \[ (x - 1)(x - 7) = 0 \] **Step 4: Apply the Zero Product Property.** Set each factor equal to zero: 1. \( x - 1 = 0 \) \[ x = 1 \] 2. \( x - 7 = 0 \) \[ x = 7 \] **Solution:** \[ x = 1 \quad \text{and} \quad x = 7 \] --- ### **Problem 2c.** \[ x^{2} - 10x - 11 = 0 \] **Step 1: Factor the quadratic equation.** We need two numbers that multiply to **-11** and add up to **-10**. - **-11 and 1** satisfy this condition because: \[ (-11) \times 1 = -11 \] \[ (-11) + 1 = -10 \] **Step 2: Rewrite the equation in its factored form.** \[ (x - 11)(x + 1) = 0 \] **Step 3: Apply the Zero Product Property.** Set each factor equal to zero: 1. \( x - 11 = 0 \) \[ x = 11 \] 2. \( x + 1 = 0 \) \[ x = -1 \] **Solution:** \[ x = 11 \quad \text{and} \quad x = -1 \] --- ### **Problem 2d.** \[ 49 - x^{2} = 0 \] **Step 1: Rewrite the equation to resemble a difference of squares.** \[ 49 - x^{2} = 0 \] \[ (7)^2 - (x)^2 = 0 \] **Step 2: Factor using the difference of squares formula:** \[ a^{2} - b^{2} = (a - b)(a + b) \] \[ (7 - x)(7 + x) = 0 \] **Step 3: Apply the Zero Product Property.** Set each factor equal to zero: 1. \( 7 - x = 0 \) \[ x = 7 \] 2. \( 7 + x = 0 \) \[ x = -7 \] **Solution:** \[ x = 7 \quad \text{and} \quad x = -7 \] --- ### **Problem 2e.** \[ (x + 4)(x + 5) - 30 = 0 \] **Step 1: Expand the equation to simplify.** \[ (x + 4)(x + 5) - 30 = 0 \] \[ x^{2} + 5x + 4x + 20 - 30 = 0 \] \[ x^{2} + 9x - 10 = 0 \] **Step 2: Factor the quadratic equation.** We need two numbers that multiply to **-10** and add up to **9**. - **10 and -1** satisfy this condition because: \[ 10 \times (-1) = -10 \] \[ 10 + (-1) = 9 \] **Step 3: Rewrite the equation in its factored form.** \[ (x + 10)(x - 1) = 0 \] **Step 4: Apply the Zero Product Property.** Set each factor equal to zero: 1. \( x + 10 = 0 \) \[ x = -10 \] 2. \( x - 1 = 0 \) \[ x = 1 \] **Solution:** \[ x = -10 \quad \text{and} \quad x = 1 \] --- **Summary of Solutions:** - **2a.** \( x = -3 \), \( x = -5 \) - **2b.** \( x = 1 \), \( x = 7 \) - **2c.** \( x = 11 \), \( x = -1 \) - **2d.** \( x = 7 \), \( x = -7 \) - **2e.** \( x = -10 \), \( x = 1 \)