Find three consecutive integers such that the product of the first and third exceeds the product of 8 and the second by 32.

Answer 1

Let integers be x, x + 1, x + 2

x(x + 2) = 8(x + 1) + 32

x^2 + 2x = 8x + 8 + 32

x^2-6x - 40 = 0

(x - 10)(x + 4) = 0

x = 10

x = -4

Calculate two sets of x + 1 and x + 2 from these.

Answer 2

The three integers:

n, n + 1, and n + 2

n(n + 2) = 8(n + 1) + 12

Can the little child solve for n and answer?