Question

$n^3+2n^2-5n-6,$ where n is any positive integer, is always divisible by

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is A 2

Let $f\left(n\right)=n^3+2n^2-5n-6$

By factorisation we get,

$f\left(n\right)=(n-2)(n+1)(n+3)$

Case 1: If n is even then, n = 2k for some integer k.

So, f (2k) = (2k – 2) (2k + 1) (2k + 3)

= 2 (k – 1) (2k + 1) (2k + 3)

Thus, f (2k) is divisible by 2.

Case 2: If n is odd then, n = 2p + 1 for some integer p.

So, f (2p + 1) = (2p + 1 – 2) (2p + 1 + 1) (2p + 1 + 3)

= (2p – 1) (2p + 2) (2p + 4)

= (2p – 1) × 2 × (p + 1) × 2 × (p + 2)

= 4 (2p – 1) (p + 1) (p + 2)

Thus, f (2p + 1) is also divisible by 2.

Hence, the correct answer is option (a)