# math

Show that a function f(n)= 2n^3+n^2+6n+3 always produces a number that is divisible by an odd number greater than 1, for any natural number, n.

2n^3+n^2+6n+3
= n^2(2n+1) + 3(2n+1)
= (2n+1)(n^2 + 3)

what's next?

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1. 2n is even, so 2n+1 is odd
since n is a natural number, 2n+1 > 1

so, the value is a multiple of an odd number. You're done.

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