Find the middle integer of 5 positive consecutive even integers whose product of the second and fifth integers is equal to 72.

Sure, let's solve this step by step:

Step 1: Let's denote the five consecutive even integers as n, n+2, n+4, n+6, and n+8.

Step 2: According to the problem, the product of the second and fifth integers is 72. So, we can write the equation as (n+2)*(n+8) = 72.

Step 3: Expanding the equation gives n^2 + 10n + 16 = 72.

Step 4: Simplifying the equation gives n^2 + 10n - 56 = 0.

Step 5: This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest method. The factors of -56 that add up to 10 are -4 and 14. So, the factored form of the equation is (n - 4)(n + 14) = 0.

Step 6: Setting each factor equal to zero gives the solutions n = 4 and n = -14. Since we're looking for positive integers, we discard n = -14.

Step 7: So, the five consecutive even integers are 4, 6, 8, 10, and 12. The middle integer of these five numbers is 8.