Note that n³ + 5n² + 2n –8 = n(n²- n) + 6(n²- n) + 8n – 8, so we can rewrite the given algebraic expression as

((n + 6) (n²- n) + 8(n – 1))/(n(n – 1))

If n= 1, then the original expression is undefined, so we may take it that n is not equal to 1, and simplify by dividing by (n – 1), to get

(n + 6) + 8/n

This will be an integer whenever n is a factor of 8 – except for n = 1 as we excluded this earlier - ie for n = 2, 4, or 8 or, for negative integers, n = -1, -2, -4, or -8.

The expression can equal any of: -3, 0, 12 or 15.

Challenge 278: Is it an Integer?