Prove for every odd integer n, 4n^2 + 3n + 2 is odd

Prove for every odd integer n, 4n^2 + 3n + 2 is odd

Let n be an odd integer, so n = 2h + 1 for some integer h.

So 4n^2 + 3n + 2 = 4(2h+1)^2 + 3(2h+1) + 2 =

16h^2 + 16h + 4 + 6h + 3 + 2 =

16h^2 + 22h + 9 =

2(8h^2 + 11h + 4) + 1, which is odd, and therefore

4n^2 + 3n + 2 is odd for every odd integer n.