find 3 consecutive odd integers such that the product of the first and the second exceeds the third by 8.

3, 5, and 7

?

Let x be the middle odd integer. Then the smaller odd integer is x-2, and the bigger is x+2

The problem requires that
x(x-2)-(x+2) = 8

or

x2 - 3x -10 = 0

Solution of the quadratic equation gives
x= -2 or x = 5
Since -2 is not an odd integer, it will be discarded.

Thus the three consecutive odd integers are
3,5,7

Check:
3*5-7=8 OK

Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8

To find the three consecutive odd integers, we can assign variables and set up equations based on the given information.

Let's assume the first odd integer as "x".
Since they are consecutive odd integers, the second odd integer would be "x + 2" and the third odd integer would be "x + 4".

According to the problem, the product of the first and the second integers exceeds the third integer by 8. Mathematically, we can express this as:

(x) * (x + 2) = (x + 4) + 8

Now, let's solve this equation to find the value of "x".

Expanding the left side of the equation:
(x^2 + 2x) = (x + 12)

Rearranging the equation and simplifying:
x^2 + 2x = x + 12
x^2 + 2x - x - 12 = 0
x^2 + x - 12 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula. In this case, the equation can be factored as:
(x + 4)(x - 3) = 0

Setting each factor to zero:
x + 4 = 0 or x - 3 = 0

Solving each equation:
x = -4 or x = 3

Since we are looking for consecutive odd integers, the solution x = 3 would be suitable.

Therefore, the three consecutive odd integers are: 3, 5, and 7.